洛谷 P1622 释放囚犯（区间dp）

题目链接

https://www.luogu.com.cn/problem/P1622

思路

$dp[i][j]$表示释放区间$[i,j]$的罪犯所需的最小的肉，状态转移方程为：

$dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k+1][j]+a[j+1]-a[i-1]-1-1)$。

其中，$k$为分断点，$a[j+1]-a[i-1]-1$表示这个区间内罪犯的人数，最后$-1$是因为不需要给释放的罪犯喂肉。

代码

#include <bits/stdc++.h>

using namespace std;

#define int long long
#define double long double

typedef long long i64;
typedef unsigned long long u64;
typedef pair<int, int> pii;

const int N = 1e2 + 5, M = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f3f3f3f3f;

int p, q;
int a[N];
int dp[N][N];
void solve()
{
    cin >> p >> q;
    for (int i = 1; i <= q; i++)
    {
        cin >> a[i];
    }
    a[0] = 0, a[q + 1] = p + 1;
    for (int len = 1; len <= q; len++)
    {
        for (int i = 1; i + len - 1 <= q; i++)
        {
            int j = i + len - 1;
            dp[i][j] = inf;
            for (int k = i; k <= j; k++)
            {
                dp[i][j] = min(dp[i][j], dp[i][k - 1] + dp[k + 1][j] + a[j + 1] - a[i - 1] - 1 - 1);
            }
        }
    }
    cout << dp[1][q] << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int test = 1;
    // cin >> test;
    for (int i = 1; i <= test; i++)
    {
        solve();
    }
    return 0;
}