洛谷 P1880 [NOI1995]石子合并（区间DP）

嗯...

题目链接：https://www.luogu.org/problem/P1880

这道题特点在于石子是一个环，所以让a[i+n] = a[i]（两倍长度）即可解决环的问题，然后注意求区间最小值的时候dp要初始化为一个很大的数...

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;

int dp1[205][205], dp2[205][205], sum[205], a[205];

int main(){
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {scanf("%d", &a[i]); a[i + n] = a[i];}
    for(int i = 1; i <= n * 2; i++) {sum[i] = sum[i - 1] + a[i];}
    for(int l = 1; l <= n; l++){
        for(int i = 1; i <= 2 * n; i++){
            int j = i + l;
            dp2[i][j] = 0x3f3f3f;// 求最小值初始化为很大 
            if(j > 2 * n) break;
            for(int k = i; k < j; k++){
                dp1[i][j] = max(dp1[i][j], dp1[i][k] + dp1[k + 1][j] + sum[j] - sum[i - 1]);
                dp2[i][j] = min(dp2[i][j], dp2[i][k] + dp2[k + 1][j] + sum[j] - sum[i - 1]);
            }
        }
    }
    int maxx = 0, minn = 0x3f3f;
    for(int i = 1; i <= n; i++){
        int j = i + n - 1;
        maxx = max(maxx, dp1[i][j]);
        minn = min(minn, dp2[i][j]);
    }
    printf("%d
%d
", minn, maxx);
    return 0;
}