第一周 1.17-1.23
1.17
CF 614 E Necklace
类似于函数图像有两条对称轴 必然是周期的。
如果循环次数是奇数 每一循环节都要是回文 把奇数的放中间 偶数放两边 （所以奇数的不止1个就无解）。
循环次数偶 随便放一个循环节 相邻的对称一下即可。
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 const int maxn = 1e5 + 10; 5 int num[26]; 6 char s[maxn]; 7 8 int gcd(int a, int b) 9 { 10 return a % b ? gcd(b, a % b) : b; 11 } 12 13 int main(void) 14 { 15 int n, cnt = 0; 16 scanf("%d", &n); 17 for(int i = 0; i < n; i++) scanf("%d", num + i); 18 int tmp = num[0]; 19 for(int i = 1; i < n; i++) tmp = gcd(tmp, num[i]); 20 for(int i = 0; i < n; i++) num[i] /= tmp; 21 if(tmp & 1) 22 { 23 int pos1 = -1; 24 for(int i = 0; i < n; i++) 25 { 26 if(num[i] & 1) 27 { 28 if(pos1 == -1) pos1 = i; 29 else 30 { 31 puts("0"); 32 for(int k = 0; k < n; k++) 33 for(int j = 0; j < tmp * num[k]; j++) 34 putchar('a' + k); 35 puts(""); 36 return 0; 37 } 38 } 39 } 40 for(int i = 0; i < n; i++) 41 { 42 if(num[i] & 1) num[i]--; 43 for(int j = 0; j < num[i] / 2; j++) 44 s[cnt++] = 'a' + i; 45 } 46 int l = cnt; 47 if(pos1 != -1) s[cnt++] = 'a' + pos1; 48 for(int j = 1; j <= l; j++) s[cnt++] = s[l-j]; 49 s[cnt] = 0; 50 printf("%d\n", tmp); 51 for(int j = 0; j < tmp; j++) printf("%s", s); 52 puts(""); 53 } 54 else 55 { 56 for(int i = 0; i < n; i++) 57 for(int j = 0; j < num[i]; j++) 58 s[cnt++] = 'a' + i; 59 for(int i = cnt; i; i--) 60 s[cnt++] = s[i-1]; 61 s[cnt] = 0; 62 printf("%d\n", tmp); 63 for(int j = 0; j < tmp / 2; j++) printf("%s", s); 64 puts(""); 65 } 66 return 0; 67 }
原来看后缀数组是上个月的事情了？
POJ 2217 Secretary
最长公共子串。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 const int maxn = 22222; 7 char s[maxn]; 8 int n, k, SA[maxn], r[maxn], tmp[maxn], h[maxn]; 9 bool cmp(int i, int j) 10 { 11 if(r[i] != r[j]) return r[i] < r[j]; 12 return ( i + k <= n ? r[i+k] : -1 ) < ( j + k <= n ? r[j+k] : -1 ); 13 } 14 void get_SA() 15 { 16 for(int i = 0; i <= n; i++) 17 { 18 SA[i] = i; 19 r[i] = i < n ? s[i] : -1; 20 } 21 for(k = 1; k <= n; k <<= 1) 22 { 23 sort(SA, SA + n + 1, cmp); 24 tmp[SA[0]] = 0; 25 for(int i = 1; i <= n; i++) tmp[SA[i]] = tmp[SA[i-1]] + cmp(SA[i-1], SA[i]); 26 memcpy(r, tmp, sizeof(r)); 27 } 28 return; 29 } 30 void get_height() 31 { 32 for(int i = 0; i <= n; i++) r[SA[i]] = i; 33 int k = 0; 34 for(int i = 0; i < n; i++) 35 { 36 if(k) k--; 37 int j = SA[r[i]-1]; 38 while(s[i+k] == s[j+k]) k++; 39 h[r[i]] = k; 40 } 41 return; 42 } 43 44 int main(void) 45 { 46 int N; 47 scanf("%d", &N); 48 getchar(); 49 while(N--) 50 { 51 gets(s); 52 int m = strlen(s); 53 s[m] = '~'; 54 gets(s + m + 1); 55 n = strlen(s); 56 get_SA(); 57 get_height(); 58 int ans = 0; 59 for(int i = 1; i <= n; i++) 60 if( ( SA[i] <= m ) != ( SA[i-1] <= m ) && h[i] > ans) 61 ans = h[i]; 62 printf("Nejdelsi spolecny retezec ma delku %d.\n", ans); 63 } 64 return 0; 65 }
UVA 11107 Life Forms
讨厌UVA。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 const int maxn = 111111; 7 int n, k, SA[maxn], r[maxn], tmp[maxn], h[maxn]; 8 int range[maxn], flag[111]; 9 char s[maxn]; 10 11 bool cmp(int i, int j) 12 { 13 if(r[i] != r[j]) return r[i] < r[j]; 14 return ( i + k <= n ? r[i+k] : -1 ) < ( j + k <= n ? r[j+k] : -1 ); 15 } 16 17 void get_SA() 18 { 19 for(int i = 0; i <= n; i++) 20 { 21 SA[i] = i; 22 r[i] = i < n ? s[i] : -1; 23 } 24 for(k = 1; k <= n; k <<= 1) 25 { 26 sort(SA, SA + n + 1, cmp); 27 tmp[SA[0]] = 0; 28 for(int i = 1; i <= n; i++) tmp[SA[i]] = tmp[SA[i-1]] + cmp(SA[i-1], SA[i]); 29 memcpy(r, tmp, sizeof(r)); 30 } 31 return; 32 } 33 34 void get_height() 35 { 36 for(int i = 0; i <= n; i++) r[SA[i]] = i; 37 int k = 0; 38 for(int i = 0; i < n; i++) 39 { 40 if(k) k--; 41 int j = SA[r[i]-1]; 42 while(s[i+k] == s[j+k]) k++; 43 h[r[i]] = k; 44 } 45 return; 46 } 47 48 int main(void) 49 { 50 int N, fir = 1; 51 while(~scanf("%d", &N) && N) 52 { 53 if(!fir) puts(""); 54 fir = 0; 55 if(N == 1) 56 { 57 scanf("%s", s); 58 printf("%s\n\n", s); 59 continue; 60 } 61 int pos = 0; 62 for(int i = 1; i <= N; i++) 63 { 64 scanf("%s", s + pos); 65 for(int j = pos; s[j]; j++) range[j] = i; 66 pos = strlen(s); 67 s[pos++] = 130 + i; 68 } 69 s[pos] = 0; 70 n = strlen(s); 71 get_SA(); 72 get_height(); 73 int l = 0, r = n, mid; 74 while(l < r) 75 { 76 int ok = 0; 77 mid = (l + r + 1) / 2; 78 for(int i = 1; i <= n; i++) 79 { 80 if(h[i] < mid) continue; 81 int cnt = 0; 82 memset(flag, 0, sizeof(flag)); 83 for(int j = i; j <= n; j++) 84 { 85 if(h[j] < mid) break; 86 int a = range[SA[j]], b = range[SA[j-1]]; 87 if(a == b) continue; 88 if(!flag[a]) cnt++; 89 if(!flag[b]) cnt++; 90 flag[a] = flag[b] = 1; 91 } 92 if(cnt > N / 2) {ok = 1;break;} 93 } 94 if(ok) l = mid; 95 else r = mid - 1; 96 } 97 if(l) for(int i = 1; i <= n; i++) 98 { 99 if(h[i] < l) continue; 100 int cnt = 0; 101 memset(flag, 0, sizeof(flag)); 102 for(int j = i; j <= n; j++) 103 { 104 if(h[j] < l) break; 105 int a = range[SA[j]], b = range[SA[j-1]]; 106 if(!a || !b || a == b) continue; 107 if(!flag[a]) cnt++; 108 if(!flag[b]) cnt++; 109 flag[a] = flag[b] = 1; 110 i = j; 111 } 112 if(cnt > N / 2) 113 { 114 for(int j = 0; j < l; j++) putchar(s[SA[i]+j]); 115 puts(""); 116 } 117 } 118 else puts("?"); 119 } 120 return 0; 121 }
POJ 1743 Musical Theme
做差搞SA。相邻要大于mid而不是大于等于mid才不重叠。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 const int maxn = 111111; 7 int n, k, SA[maxn], r[maxn], tmp[maxn], h[maxn]; 8 int s[maxn]; 9 10 bool cmp(int i, int j) 11 { 12 if(r[i] != r[j]) return r[i] < r[j]; 13 return ( i + k <= n ? r[i+k] : -1 ) < ( j + k <= n ? r[j+k] : -1 ); 14 } 15 16 void get_SA() 17 { 18 for(int i = 0; i <= n; i++) 19 { 20 SA[i] = i; 21 r[i] = i < n ? s[i] : -1; 22 } 23 for(k = 1; k <= n; k <<= 1) 24 { 25 sort(SA, SA + n + 1, cmp); 26 tmp[SA[0]] = 0; 27 for(int i = 1; i <= n; i++) tmp[SA[i]] = tmp[SA[i-1]] + cmp(SA[i-1], SA[i]); 28 memcpy(r, tmp, sizeof(r)); 29 } 30 return; 31 } 32 33 void get_height() 34 { 35 for(int i = 0; i <= n; i++) r[SA[i]] = i; 36 int k = 0; 37 for(int i = 0; i < n; i++) 38 { 39 if(k) k--; 40 int j = SA[r[i]-1]; 41 while(s[i+k] == s[j+k]) k++; 42 h[r[i]] = k; 43 } 44 return; 45 } 46 47 int main(void) 48 { 49 int N; 50 while(~scanf("%d", &N) && N) 51 { 52 for(int i = 0; i < N; i++) scanf("%d", s + i); 53 for(int i = 1; i < N; i++) s[i-1] -= s[i]; 54 n = N - 1; 55 get_SA(); 56 get_height(); 57 int l = 0, r = n - 1, mid; 58 while(l < r) 59 { 60 int ok = 0; 61 mid = (l + r + 1) / 2; 62 for(int i = 1; i < n; i++) if(h[i] >= mid) 63 { 64 int Max = SA[i-1], Min = SA[i-1]; 65 for(int j = i; j < n; j++) 66 { 67 if(h[j] < mid) break; 68 Max = max(Max, SA[j]); 69 Min = min(Min, SA[j]); 70 i = j; 71 } 72 if(Max - Min > mid) {ok = 1;break;} 73 } 74 if(ok) l = mid; 75 else r = mid - 1; 76 } 77 if(l > 3) printf("%d\n", l + 1); 78 else puts("0"); 79 } 80 return 0; 81 }
POJ 3261 Milk Patterns
上面改改就好啦。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 const int maxn = 111111; 7 int n, k, SA[maxn], r[maxn], tmp[maxn], h[maxn]; 8 int s[maxn]; 9 10 bool cmp(int i, int j) 11 { 12 if(r[i] != r[j]) return r[i] < r[j]; 13 return ( i + k <= n ? r[i+k] : -1 ) < ( j + k <= n ? r[j+k] : -1 ); 14 } 15 16 void get_SA() 17 { 18 for(int i = 0; i <= n; i++) 19 { 20 SA[i] = i; 21 r[i] = i < n ? s[i] : -1; 22 } 23 for(k = 1; k <= n; k <<= 1) 24 { 25 sort(SA, SA + n + 1, cmp); 26 tmp[SA[0]] = 0; 27 for(int i = 1; i <= n; i++) tmp[SA[i]] = tmp[SA[i-1]] + cmp(SA[i-1], SA[i]); 28 memcpy(r, tmp, sizeof(r)); 29 } 30 return; 31 } 32 33 void get_height() 34 { 35 for(int i = 0; i <= n; i++) r[SA[i]] = i; 36 int k = 0; 37 for(int i = 0; i < n; i++) 38 { 39 if(k) k--; 40 int j = SA[r[i]-1]; 41 while(s[i+k] == s[j+k]) k++; 42 h[r[i]] = k; 43 } 44 return; 45 } 46 47 int main(void) 48 { 49 int K; 50 while(~scanf("%d%d", &n, &K)) 51 { 52 for(int i = 0; i < n; i++) scanf("%d", s + i); 53 get_SA(); 54 get_height(); 55 int l = 0, r = n, mid; 56 while(l < r) 57 { 58 int ok = 0; 59 mid = (l + r + 1) / 2; 60 for(int i = 1; i <= n; i++) if(h[i] >= mid) 61 { 62 int cnt = 1; 63 for(int j = i; j <= n; j++) 64 { 65 if(h[j] < mid) break; 66 cnt++; 67 i = j; 68 } 69 if(cnt >= K) {ok = 1;break;} 70 } 71 if(ok) l = mid; 72 else r = mid - 1; 73 } 74 printf("%d\n", l); 75 } 76 return 0; 77 }
SPOJ DISUBSTR Distinct Substrings
每个后缀对答案的贡献是n - sa[i] - height[i]。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 const int maxn = 111111; 7 int n, k, SA[maxn], r[maxn], tmp[maxn], h[maxn]; 8 char s[maxn]; 9 10 bool cmp(int i, int j) 11 { 12 if(r[i] != r[j]) return r[i] < r[j]; 13 return ( i + k <= n ? r[i+k] : -1 ) < ( j + k <= n ? r[j+k] : -1 ); 14 } 15 16 void get_SA() 17 { 18 for(int i = 0; i <= n; i++) 19 { 20 SA[i] = i; 21 r[i] = i < n ? s[i] : -1; 22 } 23 for(k = 1; k <= n; k <<= 1) 24 { 25 sort(SA, SA + n + 1, cmp); 26 tmp[SA[0]] = 0; 27 for(int i = 1; i <= n; i++) tmp[SA[i]] = tmp[SA[i-1]] + cmp(SA[i-1], SA[i]); 28 memcpy(r, tmp, sizeof(r)); 29 } 30 return; 31 } 32 33 void get_height() 34 { 35 for(int i = 0; i <= n; i++) r[SA[i]] = i; 36 int k = 0; 37 for(int i = 0; i < n; i++) 38 { 39 if(k) k--; 40 int j = SA[r[i]-1]; 41 while(s[i+k] == s[j+k]) k++; 42 h[r[i]] = k; 43 } 44 return; 45 } 46 47 int main(void) 48 { 49 int T; 50 scanf("%d", &T); 51 for(int kase = 1; kase <= T; kase++) 52 { 53 scanf("%s", s); 54 n = strlen(s); 55 get_SA(); 56 get_height(); 57 int ans = n - SA[0]; 58 for(int i = 1; i <= n; i++) ans += n - SA[i] - h[i]; 59 printf("%d\n", ans); 60 } 61 return 0; 62 }
1.18
什么都没干。
1.19
POJ 3693 Maximum repetition substring
看了好多题解才看懂。然而并不好想阿。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 const int maxn = 111111; 7 int n, k, SA[maxn], r[maxn], tmp[maxn], h[maxn], rmq[maxn][20]; 8 char s[maxn]; 9 10 bool cmp(int i, int j) 11 { 12 if(r[i] != r[j]) return r[i] < r[j]; 13 return ( i + k <= n ? r[i+k] : -1 ) < ( j + k <= n ? r[j+k] : -1 ); 14 } 15 16 void get_SA() 17 { 18 for(int i = 0; i <= n; i++) 19 { 20 SA[i] = i; 21 r[i] = i < n ? s[i] : -1; 22 } 23 for(k = 1; k <= n; k <<= 1) 24 { 25 sort(SA, SA + n + 1, cmp); 26 tmp[SA[0]] = 0; 27 for(int i = 1; i <= n; i++) tmp[SA[i]] = tmp[SA[i-1]] + cmp(SA[i-1], SA[i]); 28 memcpy(r, tmp, sizeof(r)); 29 } 30 return; 31 } 32 33 void get_height() 34 { 35 for(int i = 0; i <= n; i++) r[SA[i]] = i; 36 int k = 0; 37 for(int i = 0; i < n; i++) 38 { 39 if(k) k--; 40 int j = SA[r[i]-1]; 41 while(s[i+k] == s[j+k]) k++; 42 h[r[i]] = k; 43 } 44 return; 45 } 46 47 void RMQ_init() 48 { 49 for(int i = 1; i <= n; i++) rmq[i][0] = h[i]; 50 for(int j = 1; (1 << j) <= n; j++) 51 for(int i = 1; i + ( 1 << j ) - 1 <= n; i++) 52 rmq[i][j] = min(rmq[i][j-1] , rmq[i+(1<<j-1)][j-1]); 53 } 54 55 int RMQ_query(int l, int r) 56 { 57 int k = 0; 58 while( ( 1 << (k + 1) ) <= r - l + 1 ) k++; 59 return min(rmq[l][k], rmq[r-(1<<k)+1][k]); 60 } 61 62 int lcp(int i, int j) 63 { 64 i = r[i], j = r[j]; 65 if(i > j) swap(i, j); 66 return RMQ_query(i + 1, j); 67 } 68 69 int main(void) 70 { 71 int kase = 0; 72 while(~scanf("%s", s) && s[0] != '#') 73 { 74 n = strlen(s); 75 get_SA(); 76 get_height(); 77 RMQ_init(); 78 int Max = 1, MaxL = 1; 79 for(int L = 1; L < n; L++) 80 { 81 for(int i = 0; i + L < n; i += L) 82 { 83 int cur = lcp(i, i + L), tmp = cur / L + 1; 84 int pos = i - L + cur % L; 85 if(pos >= 0 && lcp(pos, pos + L) >= tmp * L) tmp++; 86 if(tmp > Max) Max = tmp, MaxL = L; 87 } 88 } 89 for(int i = 1; i <= n; i++) 90 { 91 if(SA[i] + MaxL <= n && lcp(SA[i], SA[i] + MaxL) >= Max * MaxL - MaxL) 92 { 93 printf("Case %d: ", ++kase); 94 for(int j = 0; j < MaxL * Max; j++) putchar(s[SA[i]+j]); 95 puts(""); break; 96 } 97 } 98 } 99 return 0; 100 }
POJ 3415 Common Substrings
好烦啊。还要搞个单调栈。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 typedef long long LL; 7 const int maxn = 222222; 8 int n, m, k, K, SA[maxn], r[maxn], tmp[maxn], h[maxn]; 9 LL sta[maxn], cnt[maxn]; 10 char s[maxn]; 11 12 bool cmp(int i, int j) 13 { 14 if(r[i] != r[j]) return r[i] < r[j]; 15 return ( i + k <= n ? r[i+k] : -1 ) < ( j + k <= n ? r[j+k] : -1 ); 16 } 17 18 void get_SA() 19 { 20 for(int i = 0; i <= n; i++) 21 { 22 SA[i] = i; 23 r[i] = i < n ? s[i] : -1; 24 } 25 for(k = 1; k <= n; k <<= 1) 26 { 27 sort(SA, SA + n + 1, cmp); 28 tmp[SA[0]] = 0; 29 for(int i = 1; i <= n; i++) tmp[SA[i]] = tmp[SA[i-1]] + cmp(SA[i-1], SA[i]); 30 memcpy(r, tmp, sizeof(r)); 31 } 32 return; 33 } 34 35 void get_height() 36 { 37 for(int i = 0; i <= n; i++) r[SA[i]] = i; 38 int k = 0; 39 for(int i = 0; i < n; i++) 40 { 41 if(k) k--; 42 int j = SA[r[i]-1]; 43 while(s[i+k] == s[j+k]) k++; 44 h[r[i]] = k; 45 } 46 return; 47 } 48 49 LL cal(int flag) 50 { 51 LL ret = 0, top = 0, sum = 0; 52 for(int i = 1; i <= n; i++) 53 { 54 if(h[i] < K) top = sum = 0; 55 else 56 { 57 int tmp = 0; 58 if( (SA[i-1] < m) ^ flag ) tmp = 1, sum += (LL) h[i] - K + 1; 59 while(top && sta[top-1] > h[i]) 60 { 61 top--; 62 sum -= cnt[top] * (sta[top] - h[i]); 63 tmp += cnt[top]; 64 } 65 sta[top] = h[i], cnt[top++] = tmp; 66 if( (SA[i] > m) ^ flag ) ret += sum; 67 } 68 } 69 return ret; 70 } 71 72 int main(void) 73 { 74 while(~scanf("%d", &K) && K) 75 { 76 scanf("%s", s); 77 m = strlen(s); 78 s[m] = '~'; 79 scanf("%s", s + m + 1); 80 n = strlen(s); 81 get_SA(); 82 get_height(); 83 printf("%I64d\n",cal(0) + cal(1)); 84 } 85 return 0; 86 }
1.20
打个TC。没上黄。
TC SRM 679 DIV2 500 ContestScoreboard
FST了。就是一个模拟题。
感觉学习到了就是sstream这个玩意儿。- -
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <string> 5 #include <sstream> 6 #include <vector> 7 #include <algorithm> 8 using namespace std; 9 typedef long long LL; 10 typedef pair<int, int> pii; 11 12 struct team 13 { 14 string Name; 15 int id; 16 LL Score; 17 }T[111]; 18 19 bool cmp(team A, team B) 20 { 21 if(A.Score != B.Score) return A.Score > B.Score; 22 return A.Name < B.Name; 23 } 24 25 class ContestScoreboard 26 { 27 public: 28 vector <int> findWinner(vector <string> scores) 29 { 30 int N = scores.size(); 31 vector<pii> tmp[111]; 32 vector<int> time; 33 int ans[111]; 34 memset(ans, 0, sizeof(ans)); 35 for(int i = 0; i < N; i++) 36 { 37 char c; 38 T[i].id = i; 39 stringstream ss; 40 ss.str(scores[i]); 41 ss >> T[i].Name; 42 int Sc, Ti; 43 while( ss >> Sc >> c >> Ti) tmp[i].push_back(pii(Ti,Sc)), time.push_back(Ti + 1); 44 sort(tmp[i].begin(), tmp[i].end()); 45 } 46 time.push_back(0); 47 sort(time.begin(), time.end()); 48 for(int i = 0; i < time.size(); i++) 49 { 50 for(int j = 0; j < N; j++) 51 { 52 T[j].Score = 0LL; 53 for(int k = 0; k < tmp[T[j].id].size(); k++) 54 { 55 if(tmp[T[j].id][k].first >= time[i]) break; 56 T[j].Score += (LL) tmp[T[j].id][k].second; 57 } 58 } 59 sort(T, T + N, cmp); 60 ans[T[0].id] = 1; 61 } 62 vector<int> ret; 63 for(int i = 0; i < N; i++) ret.push_back(ans[i]); 64 return ret; 65 } 66 };
1.21
TC SRM 679 DIV2 1000 ForbiddenStreets
暴力思路很正确阿。然而没写完。
Floyd时每两点维护一个边集。又一次用了SLT的集合运算。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 #include <set> 6 #include <algorithm> 7 using namespace std; 8 int G[111][111]; 9 10 class ForbiddenStreets 11 { 12 public: 13 vector <int> find(int N, vector <int> A, vector <int> B, vector <int> time) 14 { 15 int m = A.size(); 16 memset(G, 0 ,sizeof(G)); 17 set<int> r[111][111]; 18 for(int i = 0; i < m; i++) 19 { 20 int a = A[i], b = B[i], d = time[i]; 21 G[a][b] = G[b][a] = d; 22 r[a][b].insert(i); 23 r[b][a].insert(i); 24 } 25 for(int k = 0; k < N; k++) 26 { 27 for(int i = 0; i < N; i++) 28 { 29 for(int j = 0; j < N; j++) 30 { 31 if(!G[i][k] || !G[k][j]) continue; 32 if(!G[i][j] || G[i][k] + G[k][j] < G[i][j]) 33 { 34 G[i][j] = G[i][k] + G[k][j]; 35 r[i][j].clear(); 36 set_union(r[i][k].begin(), r[i][k].end(), r[k][j].begin(), r[k][j].end(), inserter(r[i][j], r[i][j].begin())); 37 } 38 else if(G[i][k] + G[k][j] == G[i][j]) 39 { 40 set<int> a, b; 41 set_union(r[i][k].begin(), r[i][k].end(), r[k][j].begin(), r[k][j].end(), inserter(a, a.begin())); 42 set_intersection(a.begin(), a.end(), r[i][j].begin(), r[i][j].end(), inserter(b, b.begin())); 43 r[i][j] = b; 44 } 45 } 46 } 47 } 48 int cnt[2222]; 49 memset(cnt, 0, sizeof(cnt)); 50 for(int i = 0; i < N; i++) 51 for(int j = i + 1; j < N; j++) 52 for(set<int>::iterator it = r[i][j].begin(); it != r[i][j].end(); it++) cnt[*it]++; 53 vector<int> ret; 54 for(int i = 0; i < m; i++) ret.push_back(cnt[i]); 55 return ret; 56 } 57 };
1.22
CF 620 D Professor GukiZ and Two Arrays
场上比较sb。一直在贪。
swap2次的话。把a的组合丢map里，再遍历b的组合去二分a。
麻烦一点就是二分要检查两边的值，比如用lower_bound()要检查it和it--的key。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <algorithm> 5 #include <map> 6 using namespace std; 7 typedef long long LL; 8 typedef pair<int, int> pii; 9 LL a[2222], b[2222]; 10 map<LL, pii> M; 11 map<LL, pii>::iterator it; 12 13 int main(void) 14 { 15 int n, m; 16 LL sa = 0, sb = 0; 17 scanf("%d", &n); 18 for(int i = 1; i <= n; i++) scanf("%I64d", a + i), sa += a[i]; 19 scanf("%d", &m); 20 for(int i = 1; i <= m; i++) scanf("%I64d", b + i), sb += b[i]; 21 LL v = abs(sa - sb), pa = 0, pb = 0; 22 for(int i = 1; i <= n; i++) 23 for(int j = 1; j <= m; j++) 24 if(abs(sa - a[i] * 2LL - sb + b[j] * 2LL) < v) 25 v = abs(sa - a[i] * 2LL - sb + b[j] * 2LL), pa = i, pb = j; 26 if(!pa) {printf("%I64d\n0\n", v); return 0;} 27 28 for(int i = 1; i <= n; i++) 29 for(int j = i + 1; j <= n; j++) 30 M[(a[i] + a[j]) * 2LL] = pii(i, j); 31 32 LL ppa = 0, ppb = 0; 33 34 for(int i = 1; i <= m; i++) 35 { 36 for(int j = i + 1; j <= m; j++) 37 { 38 LL x = sa - sb + b[i] + b[i] + b[j] + b[j]; 39 it = M.lower_bound(x); 40 if(it != M.end() && abs((*it).first - x) < v) 41 { 42 v = abs((*it).first - x); 43 pii tmp = (*it).second; 44 pa = tmp.first, pb = i, ppa = tmp.second, ppb = j; 45 } 46 if(it != M.begin()) 47 { 48 it--; 49 if(abs((*it).first - x) >= v) continue; 50 v = abs((*it).first - x); 51 pii tmp = (*it).second; 52 pa = tmp.first, pb = i, ppa = tmp.second, ppb = j; 53 } 54 } 55 } 56 57 if(!ppa) printf("%I64d\n1\n%I64d %I64d\n", v, pa, pb); 58 else printf("%I64d\n2\n%I64d %I64d\n%I64d %I64d\n",v, pa, pb, ppa, ppb); 59 return 0; 60 }
CF 620 E New Year Tree
哭了太久没写不会写线段树了。
1 #include <iostream> 2 #include <cstdio> 3 #define lson (2 * p) 4 #define rson (2 * p + 1) 5 #define mid ( (tr[p].l + tr[p].r) / 2 ) 6 using namespace std; 7 typedef long long LL; 8 const int maxn = 4e5 + 10; 9 int cnt , h[maxn]; 10 int timer, dfn[maxn][2]; 11 int id[maxn<<1], col[maxn]; 12 13 struct edge 14 { 15 int to, pre; 16 } e[maxn<<1]; 17 18 void add(int from, int to) 19 { 20 cnt++; 21 e[cnt].pre = h[from]; 22 e[cnt].to = to; 23 h[from] = cnt; 24 } 25 26 void dfs(int p, int fa) 27 { 28 dfn[p][0] = ++timer; 29 id[timer] = p; 30 for(int i = h[p]; i; i = e[i].pre) 31 { 32 int to = e[i].to; 33 if(to == fa) continue; 34 dfs(to, p); 35 } 36 dfn[p][1] = ++timer; 37 } 38 39 struct seg_tree 40 { 41 int l, r, tag; 42 LL c; 43 }tr[maxn<<3]; 44 45 void pushdown(int p) 46 { 47 if(tr[p].tag) tr[p].c = 1LL << tr[p].tag; 48 if(tr[p].tag && tr[p].l != tr[p].r) tr[lson].tag = tr[rson].tag = tr[p].tag; 49 tr[p].tag = 0; 50 } 51 52 void pushup(int p) 53 { 54 pushdown(lson), pushdown(rson); 55 tr[p].c = tr[lson].c | tr[rson].c; 56 } 57 58 void build(int p, int l, int r) 59 { 60 tr[p].l = l, tr[p].r = r, tr[p].tag = tr[p].c = 0LL; 61 if(l < r) build(lson, l, mid), build(rson, mid + 1, r), pushup(p); 62 else if(id[l]) tr[p].c = 1LL << col[id[l]]; 63 } 64 65 void modify(int p, int l, int r, int c) 66 { 67 if(tr[p].l >= l && tr[p].r <= r) tr[p].tag = c; 68 pushdown(p); 69 if(tr[p].l >= l && tr[p].r <= r) return; 70 if(l <= mid) modify(lson, l, r, c); 71 if(r > mid) modify(rson, l, r, c); 72 pushup(p); 73 } 74 75 LL query(int p, int l, int r) 76 { 77 pushdown(p); 78 if(tr[p].l >= l && tr[p].r <= r) return tr[p].c; 79 LL ret = 0LL; 80 if(l <= mid) ret |= query(lson, l, r); 81 if(r > mid) ret |= query(rson, l, r); 82 pushup(p); 83 return ret; 84 } 85 86 int main(void) 87 { 88 int n, m; 89 scanf("%d %d", &n, &m); 90 for(int i = 1; i <= n; i++) scanf("%d", col + i); 91 for(int i = 1; i < n; i++) 92 { 93 int u, v; 94 scanf("%d %d", &u, &v); 95 add(u, v), add(v, u); 96 } 97 dfs(1, 0); 98 build(1, 1, timer); 99 for(int i = 1; i <= m; i++) 100 { 101 int op, v, c, ans = 0; 102 scanf("%d", &op); 103 if(op == 1) 104 { 105 scanf("%d %d", &v, &c); 106 modify(1, dfn[v][0], dfn[v][1], c); 107 } 108 else 109 { 110 scanf("%d", &v); 111 LL ret = query(1, dfn[v][0], dfn[v][1]); 112 while(ret) ans += ret & 1LL, ret >>= 1LL; 113 printf("%d\n", ans); 114 } 115 } 116 return 0; 117 }
1.23
HDU 5611 Baby Ming and phone number
SBSBSBSBSBSB
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 typedef long long LL; 6 int d[] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; 7 char * num[22] = { 8 "00000", "11111", "22222", "33333", "44444", "55555", "66666", "77777", "88888", "99999", 9 "01234", "12345", "23456", "34567", "45678", "56789", 10 "98765", "87654", "76543", "65432", "54321", "43210" 11 }; 12 13 bool judge(char * s) 14 { 15 for(int i = 0; i < 22; i++) if(strcmp(num[i], s + 6) == 0) return true; 16 int y = ( s[3] - '0' ) * 1000 + ( s[4] - '0' ) * 100 + ( s[5] - '0' ) * 10 + ( s[6] - '0' ); 17 int m = ( s[7] - '0' ) * 10 + ( s[8] - '0' ); 18 int da = ( s[9] - '0' ) * 10 + ( s[10] - '0' ); 19 if(y < 1980 || y > 2016) return false; 20 int leap = 0; 21 if( ( (y % 4 == 0) && (y % 100 != 0) ) || (y % 400 == 0) ) leap = 1; 22 if(m < 1 || m > 12) return false; 23 if(da >= 1 && da <= d[m] + (m == 2 && leap) ) return true; 24 return false; 25 } 26 27 int main(void) 28 { 29 int T; 30 scanf("%d", &T); 31 while(T--) 32 { 33 LL n, a, b, ans = 0; 34 scanf("%I64d%I64d%I64d", &n, &a, &b); 35 for(int i = 0; i < n; i++) 36 { 37 char ss[20]; 38 scanf("%s", ss); 39 if(judge(ss)) ans += a; 40 else ans += b; 41 } 42 printf("%I64d\n", ans); 43 } 44 return 0; 45 }