URAL(DP集）
这几天扫了一下URAL上面简单的DP
第一题 简单递推
1225. Flags
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 #define LL long long 7 LL dp[50][4]; 8 int main() 9 { 10 int i,n; 11 scanf("%d",&n); 12 dp[1][1] = 1;dp[1][2] = 1; 13 for(i =2 ; i <= n ; i++) 14 { 15 dp[i][1] += dp[i-1][2]+dp[i-2][1]; 16 dp[i][2] += dp[i-1][1]+dp[i-2][2]; 17 } 18 LL ans = dp[n][1]+dp[n][2]; 19 printf("%lld\n",ans); 20 return 0; 21 }
1031. Railway Tickets
二重循环 加点限制
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 using namespace std; 7 #define LL long long 8 #define INF 100000000000 9 LL dp[10010],ss[10010]; 10 int main() 11 { 12 LL l1,l2,l3,c1,c2,c3; 13 int i,j,n,s,e,a,b; 14 cin>>l1>>l2>>l3>>c1>>c2>>c3; 15 cin>>n; 16 cin>>a>>b; 17 for(i = 2; i <= n ; i++) 18 scanf("%lld",&ss[i]); 19 s = min(a,b); 20 e = max(a,b); 21 dp[s] = 0; 22 for(i = s+1; i <= e ; i++) 23 { 24 dp[i] = INF; 25 for(j = i-1 ; j >=s ; j--) 26 { 27 if(ss[i]-ss[j]>l3) 28 break; 29 if(ss[i]-ss[j]>l2) 30 dp[i] = min(dp[i],dp[j]+c3); 31 else if(ss[i]-ss[j]>l1) 32 dp[i] = min(dp[i],dp[j]+c2); 33 else 34 dp[i] = min(dp[i],dp[j]+c1); 35 } 36 } 37 cout<<dp[e]<<endl; 38 return 0; 39 }
1073. Square Country
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<cmath> 7 using namespace std; 8 #define INF 0xfffffff 9 #define N 60010 10 int dp[N]; 11 int main() 12 { 13 int i,j,n; 14 scanf("%d",&n); 15 for(i = 1; i <= n ; i++) 16 { 17 int x = sqrt(i); 18 dp[i] = INF; 19 if(x*x==i) 20 dp[i] = 1; 21 else 22 { 23 for(j = 1; j <= x ; j++) 24 dp[i] = min(dp[i],dp[i-j*j]+1); 25 } 26 } 27 printf("%d\n",dp[n]); 28 return 0; 29 }
1078. Segments
最长上升 加 路径 路径dfs回去就行
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 using namespace std; 7 int dp[510],path[510]; 8 struct node 9 { 10 int x,y,id; 11 }p[510]; 12 bool cmp(node a,node b) 13 { 14 if(a.y==b.y) 15 return a.x>b.x; 16 return a.y<b.y; 17 } 18 int o,g,flag; 19 void dfs(int u,int v) 20 { 21 int i; 22 if(flag) 23 return ; 24 path[v] = p[u].id; 25 if(v==dp[o]) 26 { 27 flag = 1; 28 for(i = dp[o]; i > 1 ; i--) 29 printf("%d ",path[i]); 30 printf("%d\n",path[1]); 31 return ; 32 } 33 for(i = 1; i < u ; i++) 34 if(p[i].x>p[u].x&&p[u].y>p[i].y) 35 { 36 if(dp[u]==dp[i]+1) 37 { 38 dfs(i,v+1); 39 } 40 } 41 } 42 int main() 43 { 44 int i,j,n; 45 scanf("%d",&n); 46 for(i = 1; i <= n ;i++) 47 { 48 scanf("%d%d",&p[i].x,&p[i].y); 49 p[i].id = i; 50 } 51 sort(p+1,p+n+1,cmp); 52 for(i = 1; i <= n ; i++) 53 { 54 dp[i] = 1; 55 for(j = 1; j < i ; j++) 56 if(p[j].x>p[i].x&&p[i].y>p[j].y) 57 dp[i] = max(dp[i],dp[j]+1); 58 } 59 int ans = 0; 60 for(i = 1; i <= n ; i++) 61 { 62 if(dp[i]>ans) 63 { 64 o = i; 65 ans = dp[i]; 66 } 67 } 68 path[++g] = p[o].id; 69 printf("%d\n",dp[o]); 70 dfs(o,1); 71 return 0; 72 }
1081. Binary Lexicographic Sequence
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 using namespace std; 7 #define LL long long 8 LL dp[50][2],sum[50],s; 9 int p[50]; 10 void init() 11 { 12 int i; 13 dp[1][0] = 1; 14 dp[1][1] = 1; 15 sum[1] = 2; 16 for(i = 2; i <=43 ; i++) 17 { 18 dp[i][0] = dp[i-1][0]+dp[i-1][1]; 19 dp[i][1] = dp[i-1][0]; 20 sum[i] = dp[i][0]+dp[i][1]; 21 } 22 } 23 int main() 24 { 25 int i,j; 26 init(); 27 int n; 28 scanf("%d%lld",&n,&s); 29 if(s>sum[n]) 30 printf("-1\n"); 31 else 32 { 33 LL ss = s; 34 while(1) 35 { 36 for(i = n ; i >= 1 ; i--) 37 if(sum[i]<=ss) 38 { 39 if(sum[i]==ss) 40 { 41 for(j = i ; j>=1 ; j-=2) 42 p[j] = 1; 43 } 44 else 45 { 46 p[i+1] = 1; 47 } 48 ss-=sum[i]; 49 break; 50 } 51 if(i==0) 52 break; 53 if(ss==0) 54 break; 55 } 56 for(i = n; i >= 1 ; i--) 57 if(p[i]) 58 printf("1"); 59 else 60 printf("0"); 61 puts(""); 62 } 63 return 0; 64 }
1119. Metro
简单DP
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<cmath> 7 using namespace std; 8 #define INF 0xfffffff 9 double dp[1010][1010]; 10 int w[1010][1010]; 11 int main() 12 { 13 int i,j,n,m,k,u,v; 14 scanf("%d%d",&n,&m); 15 scanf("%d",&k); 16 while(k--) 17 { 18 scanf("%d%d",&u,&v); 19 w[u][v] = 1; 20 } 21 for(i = 0; i <= n ; i++) 22 for(j = 0 ;j <= m ; j++) 23 dp[i][j] = INF; 24 dp[0][0] = 0; 25 for(i = 0; i <= n; i++) 26 for(j = 0; j <= m ; j++) 27 { 28 if(w[i][j]&&i>0&&j>0) 29 dp[i][j] = min(dp[i][j],dp[i-1][j-1]+100*sqrt(2.0)); 30 if(i>0) 31 dp[i][j] = min(dp[i][j],dp[i-1][j]+100.0); 32 if(j>0) 33 dp[i][j] = min(dp[i][j],dp[i][j-1]+100.0); 34 } 35 int x = dp[n][m]+0.5; 36 printf("%d\n",x); 37 return 0; 38 }
1152. False Mirrors
状压 好像写复杂了 时间跑得挺多 2s的限制 跑了1.1
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 using namespace std; 7 #define INF 0xfffffff 8 int dp[2][(1<<20)],s,a[21]; 9 int q[2][(1<<20)],p[101],sd[(1<<20)]; 10 bool f[21][(1<<20)]; 11 int main() 12 { 13 int i,j,n,o=0,g; 14 scanf("%d",&n); 15 for(j = 0 ; j < (1<<n) ; j++) 16 { 17 dp[1][j] = INF; 18 dp[0][j] = INF; 19 } 20 for(i = 1; i <= n ; i++) 21 { 22 scanf("%d",&a[i]); 23 s+=a[i]; 24 } 25 for(i = 0 ; i < (1<<n) ; i++) 26 { 27 int ss=0; 28 for(j = 0 ; j < n ; j++) 29 if(i&(1<<j)) 30 ss+=a[j+1]; 31 ss = s-ss; 32 sd[i] = ss; 33 } 34 for(i = 0 ; i < n ; i++) 35 { 36 o++; 37 p[o] = (1<<i);q[1][o] = p[o]; 38 o++; 39 p[o] = (1<<i)+(1<<(i+1)%n);q[1][o] = p[o]; 40 o++; 41 p[o] = (1<<i)+(1<<(i+1)%n)+(1<<(i+2)%n);q[1][o] = p[o]; 42 } 43 for(i = 1 ; i <= o ; i++) 44 { 45 dp[1][p[i]] = sd[p[i]]; 46 } 47 int w = o;int ans = dp[1][(1<<n)-1]; 48 for(i = 2; i <= n ; i++) 49 { 50 int tt=0; 51 for(j = 1; j <= o ; j++) 52 { 53 if(dp[(i-1)%2][q[(i-1)%2][j]]>ans) 54 continue; 55 for(g = 1 ; g <= w ; g++) 56 { 57 if(q[(i-1)%2][j]&p[g]) 58 continue; 59 int x = q[(i-1)%2][j]+p[g]; 60 dp[i%2][x] = min(dp[i%2][x],dp[(i-1)%2][q[(i-1)%2][j]]+sd[x]); 61 if(!f[i][x]) 62 { 63 f[i][x] = 1; 64 tt++; 65 q[i%2][tt] = x; 66 } 67 ans = min(ans,dp[i%2][(1<<n)-1]); 68 } 69 } 70 o = tt; 71 } 72 printf("%d\n",ans); 73 return 0; 74 }
1167. Bicolored Horses
简单DP
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 using namespace std; 7 #define INF 0xfffffff 8 int s[510],a[510],dp[510][510]; 9 int main() 10 { 11 int i,j,n,k; 12 scanf("%d%d",&n,&k); 13 for(i = 1; i <= n ; i++) 14 { 15 scanf("%d",&a[i]); 16 s[i]+=s[i-1]+a[i]; 17 } 18 for(i = 1; i <= k ;i++) 19 for(j =1 ; j <= n ; j++) 20 dp[i][j] = INF; 21 for(i = 1; i <= n-(k-1) ; i++) 22 dp[1][i] = s[i]*(i-s[i]); 23 for(i = 2; i <= k ; i++) 24 { 25 for(j = i ; j <= n-(k-i) ; j++) 26 { 27 dp[i][j] = INF; 28 for(int g = i-1 ; g < j ; g++) 29 { 30 int o = s[j]-s[g]; 31 dp[i][j] = min(dp[i][j],dp[i-1][g]+o*(j-g-o)); 32 } 33 } 34 } 35 printf("%d\n",dp[k][n]); 36 return 0; 37 }
1203. Scientific Conference
以前贪心可过的题 数据加大后DP O（n)可过
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 using namespace std; 7 int dp[30010],w[30010]; 8 struct node 9 { 10 int s,e; 11 }p[100010]; 12 bool cmp(node a,node b) 13 { 14 if(a.e==b.e) 15 return a.s<b.s; 16 return a.e<b.e; 17 } 18 int main() 19 { 20 int i,n; 21 scanf("%d",&n); 22 for(i = 1; i <= n ; i++) 23 scanf("%d%d",&p[i].s,&p[i].e); 24 sort(p+1,p+n+1,cmp); 25 for(i= 1; i <= n ; i++) 26 { 27 dp[p[i].e] = 1; 28 w[p[i].e] = p[i].s; 29 } 30 int maxz=p[n].e; 31 for(i = 1; i <= maxz ; i++) 32 { 33 if(w[i]) 34 dp[i] = max(dp[i],dp[w[i]-1]+1); 35 dp[i] = max(dp[i],dp[i-1]); 36 } 37 printf("%d\n",dp[maxz]); 38 return 0; 39 }
1260. Nudnik Photographer
这题。。以前做过 又卡了我一次
分三种情况 dp[n] = dp[n-1]+dp[n-3]+1;
12。。。。。dp[n-1]
1324.........dp[n-3]
13579.....8642 这种就一种情况
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 using namespace std; 7 #define LL long long 8 LL dp[60]; 9 int main() 10 { 11 int i,n; 12 scanf("%d",&n); 13 dp[1] =1; 14 dp[2]= 1; 15 dp[3] = 2; 16 for(i = 4 ; i <= n ; i++) 17 dp[i] = dp[i-1]+dp[i-3]+1; 18 printf("%lld\n",dp[n]); 19 return 0; 20 }
1303. Minimal Coverage
以m进行递推 因为没解的时候 还输出路径了 RE了N久。。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 using namespace std; 7 #define INF 0xfffffff 8 int dp[50100],f[50010],q[50010]; 9 int path[50100]; 10 struct node 11 { 12 int sx,x,y; 13 }p[100010]; 14 int o; 15 int main() 16 { 17 int i,j,m,x,y; 18 scanf("%d",&m); 19 memset(f,-1,sizeof(f)); 20 for(i = 1 ; i <= m ; i++) 21 dp[i] = INF; 22 dp[0] = 0; 23 while(scanf("%d%d",&x,&y)!=EOF) 24 { 25 if(x==0&&y==0) 26 break; 27 if(y>0&&x<m) 28 { 29 o++; 30 p[o].sx = x; 31 p[o].y = y; 32 if(x<0) 33 x=0; 34 p[o].x = x; 35 if(f[y]==-1||f[y]>x) 36 { 37 f[y] = x; 38 q[y] = o; 39 } 40 } 41 } 42 for(i = 1; i <= m ; i++) 43 { 44 if(f[i]!=-1) 45 { 46 if(dp[i]>dp[f[i]]+1) 47 { 48 dp[i] = dp[f[i]]+1; 49 path[dp[i]] = q[i]; 50 for(j = f[i] ; j <= i ; j++) 51 dp[j] = min(dp[j],dp[i]); 52 } 53 } 54 } 55 for(i = 1 ; i <= o ; i++) 56 { 57 if(p[i].x<m&&p[i].y>m) 58 { 59 if(dp[m]>dp[p[i].x]+1) 60 { 61 dp[m] = dp[p[i].x]+1; 62 path[dp[m]] = i; 63 } 64 } 65 } 66 if(dp[m]==INF) 67 puts("No solution"); 68 else 69 { 70 printf("%d\n",dp[m]); 71 for(i = 1; i <= dp[m] ; i++) 72 printf("%d %d\n",p[path[i]].sx,p[path[i]].y); 73 } 74 return 0; 75 }
1353. Milliard Vasya's Function
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 using namespace std; 7 #define LL long long 8 LL dp[12][90]; 9 int main() 10 { 11 int i,j,g,s; 12 scanf("%d",&s); 13 for(int p = 1; p <= 9 ; p++) 14 dp[1][p] = 1; 15 LL ans=dp[1][s]; 16 for(i = 2; i <= 9 ; i++) 17 { 18 memset(dp,0,sizeof(dp)); 19 for(int p = 1; p <= 9 ; p++) 20 dp[1][p] = 1; 21 for(int o = 2 ; o <= i ; o++) 22 for(j = 0 ; j <= s ; j++) 23 { 24 if(o==10) 25 { 26 dp[o][j]+=dp[o-1][j-1]; 27 continue; 28 } 29 for(g = 0 ; g <= 9 ; g++) 30 { 31 if(g>j) 32 continue; 33 dp[o][j] += dp[o-1][j-g]; 34 } 35 } 36 ans+=dp[i][s]; 37 } 38 if(s==1) 39 ans++; 40 printf("%lld\n",ans); 41 return 0; 42 }
1586. Threeprime Numbers
这题。。初始化数组里面一个顺序写反了 纠结了半天 三维保存两位的状态
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 using namespace std; 7 #define LL long long 8 #define mod 1000000009 9 LL dp[10010][12][12]; 10 int p[1010]; 11 void init() 12 { 13 int i,j,x=0; 14 for(i = 3; i < 1000 ; i++) 15 { 16 for(j = 2 ;j < i ; j++) 17 if(i%j==0) 18 break; 19 if(j==i) 20 { 21 p[i] = 1; 22 } 23 } 24 } 25 int main() 26 { 27 int i,j,n,o,g; 28 init(); 29 scanf("%d",&n); 30 for(i = 100 ; i <= 999 ; i++) 31 { 32 if(p[i]) 33 { 34 dp[3][(i/10)%10][i/100]++; 35 } 36 } 37 for(i = 4 ; i <= n ;i++) 38 { 39 for(j = 1 ; j <= 9 ; j++) 40 { 41 for(g = 0 ; g <= 9 ; g++) 42 { 43 for(o = 0 ; o <= 9 ; o++) 44 { 45 int x1 = j*100+g*10+o; 46 if(p[x1]) 47 { 48 dp[i][g][j] = (dp[i][g][j]+dp[i-1][o][g])%mod; 49 } 50 } 51 } 52 } 53 } 54 LL ans=0; 55 for(j = 1 ; j <= 9 ; j++) 56 for(i = 0; i <= 9 ; i++) 57 { 58 ans = (ans+dp[n][i][j])%mod; 59 } 60 printf("%lld\n",ans); 61 return 0; 62 }