#6：年兽礼包——6

石子合并，入门区间dp

 1 #include 
 2 #include 
 3 #define maxn 305
 4 #define min(a, b) a < b ? a : b
 5 
 6 int n;
 7 int sum[maxn];
 8 int f[maxn][maxn];
 9 
10 int main() {
11     memset(f, 0x3f, sizeof(f));
12     scanf("%d", &n);
13     for (int i = 1; i <= n; i++) {
14         scanf("%d", &sum[i]);
15         sum[i] += sum[i-1]; 
16         f[i][i] = 0;
17     }
18     for (int len = 2; len <= n; len++)
19         for (int l = 1; l <= n-len+1; l++) {
20             int r = l + len - 1;
21             for (int k = l; k < r; k++)
22                 f[l][r] = min(f[l][r], f[l][k] + f[k+1][r]);
23             f[l][r] += sum[r] - sum[l-1];
24         }
25     printf("%d\n", f[1][n]);
26     return 0;
27 }

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BZOJ3229，还是石子合并，但是数据量大，换算法了。100ms以内的那个实现真是厉害！我这个是朴素的……

 1 #include 
 2 #define maxn 40005
 3 #define INF 0x7fffffff
 4 #define R(a) scanf("%d", &a)
 5 #define swap(a, b) a^=b^=a^=b
 6 #define max(a, b) a > b ? a : b
 7 
 8 int n, a[maxn], ans, k, st = 1;
 9 
10 int main() {
11     R(n);
12     for (int i = 1; i <= n; i++)
13         R(a[i]);
14     a[0] = a[n+1] = INF;
15 
16     for (int m = n, i = 1; i < n; i++, m--) {
17         for (int j = st; j <= m; j++)
18             if (a[j-1] <= a[j+1]) {
19                 k = j;
20                 break;
21             }
22         
23         a[k-1] += a[k];
24         ans += a[k-1];
25 
26         for (int j = k; j <= m; j++)
27             a[j] = a[j+1];
28 
29         st = k;
30         for(k--; a[k-1] <= a[k]; swap(a[k-1], a[k]), k--);
31 
32         if (k > 2 && a[k-2] <= a[k])    st = k-1;
33         else    st = max(st-2, 1);
34     }
35 
36     printf("%d\n", ans);
37     return 0;
38 }

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POJ1179，环形拆链，乘法有负数没法直接dp，石子合并的基础上再维护一个最小值即可。

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 using namespace std;
 6 
 7 int n, ans = 0xcfcfcfcf;
 8 int Digit[105];
 9 bool signal[105];
10 int f[105][105], g[105][105];
11 int record[55], t;
12 
13 inline int cal(int a, int b, int k) {
14     return signal[k] ? a+b : a*b;
15 }
16 
17 int main() {
18     memset(f, 0xcf, sizeof(f));
19     memset(g, 0x3f, sizeof(g));
20     cin >> n;
21     for (int i = 1; i <= n; i++) {
22         char ch[3];
23         scanf("%s%d", ch, &Digit[i]);
24         signal[i-1] = signal[i-1+n] = ch[0] == 't' ? true : false;
25         f[i][i] = f[i+n][i+n] = g[i][i] = g[i+n][i+n] = Digit[i+n] = Digit[i];
26     }
27 
28     for (int len = 2; len <= n; len++)
29         for (int l = 1; l <= 2*n - len + 1; l++) {
30             int r = l + len - 1;
31             for (int k = l; k < r; k++) {
32                 f[l][r] = max(f[l][r], cal(f[l][k], f[k+1][r], k));
33                 g[l][r] = min(g[l][r], cal(g[l][k], g[k+1][r], k));
34                 if (!signal[k])    {
35                     f[l][r] = max(f[l][r], cal(g[l][k], g[k+1][r], k));
36                     g[l][r] = min(g[l][r], min(cal(f[l][k], g[k+1][r], k), cal(g[l][k], f[k+1][r], k)));
37                 }
38             }
39         }
40     
41     for (int i = 1; i <= n; i++)
42         if (ans <= f[i][i+n-1]) {
43             if (ans < f[i][i+n-1])    t = 0;
44             ans = f[i][i+n-1];
45             record[++t] = i;
46         }
47 
48     cout << ans << endl;
49     for (int i = 1; i <= t; i++)
50         printf("%d%c", record[i], " \n"[i==t]);
51 
52     return 0;
53 }

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CH金字塔，将dfs序和区间联系起来。

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 #define ll long long
 6 #define maxn 305
 7 #define mod 1000000000
 8 using namespace std;
 9 
10 char S[maxn];
11 ll f[maxn][maxn];
12 
13 int solve(int l, int r) {
14     if (~f[l][r])    return f[l][r];
15     if (l == r)    return f[l][r] = 1;
16     if (S[l-1] != S[r-1])    return f[l][r] = 0;
17     
18     f[l][r] = 0;
19     for (int k = l+1; k < r; k++)
20         f[l][r] = (f[l][r] + (ll)solve(l+1, k) * solve(k+1, r)) % mod;
21     return f[l][r];
22 }
23 
24 int main() {
25     cin >> S;
26     memset(f, -1, sizeof(f));
27     printf("%d\n", solve(1, strlen(S)));
28     return 0;
29 }

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Joyoi没有上司的舞会，入门树形dp。

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 #include 
 6 #define maxn 6005
 7 #define pb push_back
 8 using namespace std;
 9 
10 int n, H[maxn], f[maxn];
11 int m[maxn][2];
12 vector<int> v[maxn];
13 
14 int dp(int cur,int select) {
15     if (m[cur][select] != 0xcfcfcfcf)    return m[cur][select];
16 
17     if (select) {
18         m[cur][select] = H[cur];
19         for (auto i : v[cur])
20             m[cur][select] += dp(i, 0);
21     } else {
22         m[cur][select] = 0;
23         for (auto i : v[cur])
24             m[cur][select] += max(dp(i, 0), dp(i, 1));
25     }
26 
27     return m[cur][select];
28 }
29 
30 int main() {
31     cin >> n;
32     for (int i = 1; i <= n; i++)
33         cin >> H[i];
34 
35     int so, fa;
36     while (cin >> so >> fa && so) {
37         f[so] = fa;
38         v[fa].pb(so);
39     }
40     for (fa = 1; f[fa]; fa = f[fa]);
41 
42     memset(m, 0xcf, sizeof(m));
43     printf("%d\n", max(dp(fa, 1), dp(fa, 0)));
44     return 0;
45 }

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Joyoi选课，树上分组背包，果然j从0~t也是对的，没太懂书上非要倒序的意思qwq

 1 #include 
 2 #define maxn 305
 3 #define pb push_back
 4 using namespace std;
 5 
 6 int n, m, score[maxn];
 7 int f[maxn][maxn];
 8 std::vector<int> v[maxn];
 9 
10 int dp(int cur) {
11     f[cur][0] = 0;
12     for (auto i : v[cur]) {
13         dp(i);
14         for (int t = m; ~t; t--)
15             for (int j = 0; j <= t; j++)
16                 f[cur][t] = max(f[cur][t], f[cur][t-j] + f[i][j]);
17     }
18     if (cur)
19         for (int t = m; t; t--)
20             f[cur][t] = f[cur][t-1] + score[cur];
21 }
22 
23 int main() {
24     cin >> n >> m;
25     for (int i = 1; i <= n; i++) {
26         int a;
27         cin >> a >> score[i];
28         v[a].pb(i);
29     }
30     memset(f, 0xcf, sizeof(f));
31     dp(0);
32     printf("%d\n", f[0][m]);
33     return 0;
34 }

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转载于:https://www.cnblogs.com/AlphaWA/p/10351651.html