题目

一般DP代码（注意，这里只能向外推(起始状态是f(1,0)，不能向内推（不然会导致之前的羊圈被割裂））

#include <bits/stdc++.h>
using namespace std;const int MAX_N = 210;
const int MAX_M = 16;int n, m;
double p[MAX_N];
int l[MAX_M];
double dp[MAX_N][1 << MAX_M];int main() {cin >> m >> n;for (int i = 1; i <= m; i++) cin >> l[i];for (int i = 1; i <= n; i++) cin >> p[i];int mask = 1 << m;for (int i = 0; i <= n + 1; i++) {for (int j = 0; j < mask; j++) {dp[i][j] = 1e18;}}dp[1][0] = 0;for (int u = 1; u <= n; u++) {for (int s = 0; s < mask; s++) {if (dp[u][s] == 1e18) continue;// 不覆盖当前羊uif (u + 1 <= n + 1) {dp[u + 1][s] = min(dp[u + 1][s], dp[u][s] + p[u]);}// 尝试用未使用的羊圈i覆盖for (int i = 1; i <= m; i++) {if (!(s & (1 << (i - 1))) && u + l[i] - 1 <= n) {int new_s = s | (1 << (i - 1));dp[u + l[i]][new_s] = min(dp[u + l[i]][new_s], dp[u][s]);}}}}double ans = 1e18;for (int s = 0; s < mask; s++) {ans = min(ans, dp[n + 1][s]);}printf("%.2lf\n", ans);return 0;
}

记忆化搜索DP代码

#include <bits/stdc++.h>
using namespace std;
using db = double;int n, m;
db p[210];
int w[20];
db dp[210][(1 << 15) + 10];db f(int u, int s)
{if(u >= n+1) return 0;if(dp[u][s] != -1) return dp[u][s];db ret = 1e18;//不用，状态不变，但是值要增加（这里的值指的是逃跑期望）ret = p[u] + f(u+1, s);//用for(int i = 1; i <= m; i++){if(!(s & (1 << (i-1))) && u + w[i] - 1 <= n){ret = min(ret, f(u+w[i], s | (1 << (i-1))));}}return dp[u][s] = ret;
}
int main()
{cin >> m >> n;for(int i = 1; i <= m; i++)cin >> w[i];for(int i = 1; i <= n; i++)cin >> p[i];for(int i = 1; i <= n; i++)	for(int j = 0; j < 1 << m; j++)dp[i][j] = -1;printf("%.2lf", f(1, 0));
}

感想

另外勘误

这题很多解法都是没看到“最多”吗

为什么要加这个限制？

#include <bits/stdc++.h>
using namespace std;
using db = double;int n, m;
db p[210];
int w[20];
db dp[210][(1 << 15) + 10];db f(int u, int s)
{if(u >= n+1) return 0;if(dp[u][s] != -1) return dp[u][s];db ret = 1e18;//不用，状态不变，但是值要增加（这里的值指的是逃跑期望）ret = p[u] + f(u+1, s);//用for(int i = 1; i <= m; i++){if(!(s & (1 << (i-1)))){ret = min(ret, f(u+w[i], s | (1 << (i-1))));}}return dp[u][s] = ret;
}
int main()
{cin >> m >> n;for(int i = 1; i <= m; i++)cin >> w[i];for(int i = 1; i <= n; i++)cin >> p[i];for(int i = 1; i <= n; i++)	for(int j = 0; j < 1 << m; j++)dp[i][j] = -1;printf("%.2lf", f(1, 0));
}