动态规划 List

例题

    #A 传纸条（Accepted）
    #B 乘积最大 (Unaccepted)
    #C 石子合并 (Accepted)
    #D 加分二叉树 (Unaccepted)
    #E 没有上司的舞会(Unaccepted)
    #F 选课 (Accepted)
    #G 警卫安排 (Unaccepted)
    #H 通向自由的钥匙 (Unaccepted)

    #I 导弹拦截 (Unaccepted)
    #J [HAOI2010]最长公共子序列 (Unaccepted)
    #K 排列LCS问题 (Unaccepted)
    #L 尼克的任务（Accepted）
    #M 多米诺骨牌（Accepted）
    #N 最大子树和(Unaccepted)
    #O “访问”美术馆(Unaccepted)
    #P 石子合并（Accepted）
    #Q 关路灯(Unaccepted)
    #R M国王(Unaccepted)
    #S 愤怒的小鸟(Unaccepted)

---

1. LCS 最长公共子序列

/* LCS
 * Au: GG
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1005;
int n, m, d[N][N], A[N], B[N];

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
    for (int i = 1; i <= m; i++) scanf("%d", &B[i]);

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            d[i][j] = max(d[i - 1][j], d[i][j - 1]);
            if (A[i] == B[j])
                d[i][j] = max(d[i][j], d[i - 1][j - 1] + 1);
        }
    }

    printf("%d
", d[n][m]);
    return 0;
}

 2. LIS 最长上升自序列

/**
 * LIS
 * Au: GG
 **/

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 1000000 + 3;
int n, a[N], d[N];
int ans;

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
    }
    
    for (int i = 1; i <= n; i++) {
        d[i] = 1;
        for (int j = 1; j < i; j++) {
            if (a[i] > a[j] && d[j] + 1 > d[i])
                d[i] = d[j] + 1;
        }
        ans = max(ans, d[i]);
    }
    
    printf("%d
", ans);
    return 0;
}

 3. 01 背包

// 01 Knapsack
// Au: GG

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;

int n, v, w[33], f[33][20003];

int main() {
    scanf("%d%d", &v, &n);
    for (int i = 1; i <= n; i++) scanf("%d", &w[i]);

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= v; j++) {
            if (j - w[i] < 0) f[i][j] = f[i - 1][j];
            else f[i][j] = max(f[i - 1][j - w[i]] + w[i], f[i - 1][j]);
        }
    }

    printf("%d", v - f[n][v]);

    return 0;
}

 4. 完全背包

/**
 * Luogu P1616 疯狂的采药
 * Au: GG
 **/

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int n, m, d[100000+4], w[100000+3], v[100000+3];

int main() {
    scanf("%d%d", &m, &n);
    for (int i = 1; i <= n; i++) scanf("%d%d", &v[i], &w[i]);
    
    for (int i = 1; i <= n; i++) {
        for (int j = v[i]; j <= m; j++) {
            d[j] = max(d[j], d[j - v[i]] + w[i]);
        }
    }
    
    printf("%d
", d[m]);
    return 0;
}

 5. 多维背包

/**
 * Luogu P1855 榨取kkksc03
 * Au: GG
 **/

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 100 + 3, M = 200 + 3;
int n, m, t, time[N], w[N], d[N][M][M];

int main() {
    scanf("%d%d%d", &n, &m, &t);
    for (int i = 1; i <= n; i++) scanf("%d%d", &time[i], &w[i]);
    
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= m; j++) {
            for (int k = 0; k <= t; k++) {
                d[i][j][k] = d[i - 1][j][k];
                if (j - time[i] >= 0 && k - w[i] >= 0 && d[i - 1][j - time[i]][k - w[i]] + 1 > d[i][j][k])
                    d[i][j][k] = d[i - 1][j - time[i]][k - w[i]] + 1;
            }
        }
    }
    
    printf("%d
", d[n][m][t]);    
    return 0;
}

6. 树形 DP （Unaccepted）

7. 区间 DP

/* Luogu P1880 石子合并
 * Au: GG
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100 + 5;
const int inf = 2147483647;
int n, d[2 * N][2 * N], e[2 * N][2 * N], a[N], sum[2 * N];
int ans1 = inf, ans2 = - inf;

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    
    for (int i = 1; i <= 2 * n; i++) sum[i] = sum[i - 1] + a[i > n ? i % n : i];
    
    for (int i = 1; i <= 2 * n; i++) d[i][i] = 0;
    for (int len = 2; len <= n; len++) {
        for (int i = 1; i + len - 1 <= 2 * n; i++) {
            int j = i + len - 1;
            d[i][j] = inf;
            for (int k = i; k < j; k++) {
                d[i][j] = min(d[i][j], d[i][k] + d[k + 1][j] + sum[j] - sum[i - 1]);
            }
        }
    }
    for (int i = 1; i <= n; i++) ans1 = min(ans1, d[i][i + n - 1]);

    for (int i = 1; i <= 2 * n; i++) e[i][i] = 0;
    for (int len = 2; len <= n; len++) {
        for (int i = 1; i + len - 1 < 2 * n; i++) {
            int j = i + len - 1;
            e[i][j] = - inf;
            for (int k = i; k < j; k++) {
                e[i][j] = max(e[i][j], e[i][k] + e[k + 1][j] + sum[j] - sum[i - 1]);
            }
        }
    }
    
    for (int i = 1; i <= n; i++) ans2 = max(ans2, e[i][i + n - 1]);
    
    printf("%d
%d
", ans1, ans2);
    
    return 0;
}

 8. 状态压缩 DP （Unaccepted）