hdu 4597 Play Game 区间dp

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题意：

有两排数，AB依次拿，每次只能从第一/二排最左边和最右边拿

问你A拿的和是多少，假设两个人都是很聪明的

思路：

http://blog.csdn.net/shuangde800/article/details/10277697

 

出现聪明这个词的时候，这种题不是博弈论就是dp吧

dp[x][y][i][j]表示当前玩家从a堆的x~y,b堆的i~j能获得的最大价值   

区间dp

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll a[25],b[25],dp[25][25][25][25];

ll solve(int l1,int r1,int l2,int r2){
    if(dp[l1][r1][l2][r2]!=-1) return dp[l1][r1][l2][r2];

    if(l1>r1) dp[l1][r1][l2][r2] = 0;
    if(l2>r2) dp[l1][r1][l2][r2] = 0;

    ll sum = 0,ans = 0;
    if(l1<=r1)
        sum += a[r1]-a[l1-1];
    if(l2<=r2)
        sum += b[r2]-b[l2-1];

    if(l1<=r1){
        ans = max(ans,sum-solve(l1+1,r1,l2,r2));
        ans = max(ans,sum-solve(l1,r1-1,l2,r2));
    }
    if(l2<=r2){
        ans = max(ans,sum-solve(l1,r1,l2+1,r2));
        ans = max(ans,sum-solve(l1,r1,l2,r2-1));
    }
    return dp[l1][r1][l2][r2] = ans;
}

int main(){
    int T; scanf("%d",&T);
    while(T--){
        memset(dp,-1,sizeof(dp));
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        int n; scanf("%d",&n);
        for(int i=1; i<=n; i++){
            scanf("%d",&a[i]);
            a[i] += a[i-1];
        }
        for(int i=1; i<=n; i++){
            scanf("%d",&b[i]);
            b[i] += b[i-1];
        }

        cout << solve(1,n,1,n) << endl;
    }
}