HDU - 3506 Monkey Party

HDU - 3506

思路:

平行四边形不等式优化dp

这不就是石子归并(雾

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
#define LD long double
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<int, pii>
#define pdd pair<long double, long double>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head

const int N = 2e3 + 5;
int a[N], sum[N], n;
int dp[N][N], s[N][N];
int main() {
    while(~scanf("%d", &n)) {
        for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), a[i+n] = a[i];
        for (int i = 1; i <= 2*n; ++i) sum[i] = sum[i-1] + a[i];
        for (int i = 2*n; i >= 1; --i) {
            dp[i][i] = 0;
            s[i][i] = i;
            for (int j = i+1; j <= 2*n; ++j) {
                dp[i][j] = 0x7f7f7f7f;
                for (int k = s[i][j-1]; k <= s[i+1][j]; ++k) {
                    if(k+1 <= j && dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1] < dp[i][j]) {
                        dp[i][j] = dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1];
                        s[i][j] = k;
                    }
                }
            }
        }
        int ans = dp[1][n];
        for (int i = 2; i+n-1 <= 2*n; ++i) ans = min(ans, dp[i][i+n-1]);
        printf("%d\n", ans);
    }
    return 0;
}

 

posted @ 2019-05-31 13:33  Wisdom+.+  阅读(202)  评论(0编辑  收藏  举报