lightoj 1031(dp)

题意：一排数字两个人轮流从两边拿，都采用最优策略。问先手最多能拿到多少个。

思路：dp[i][j]表示先手在i-j中最多能拿到的数字之和，然后从小到大枚举区间

状态转移：

dp[j][i+j] = max(dp[j][i+j], sum[k-1] - sum[j-1] - dp[k][i+j]);
dp[j][i+j] = max(dp[j][i+j], sum[i+j] - sum[k] - dp[j][k]);

代码如下：

/**************************************************
 * Author     : xiaohao Z
 * Blog     : http://www.cnblogs.com/shu-xiaohao/
 * Last modified : 2014-05-06 11:34
 * Filename     : L_1031.cpp
 * Description     : 
 * ************************************************/
 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<unsigned int,unsigned int> puu;
typedef pair<int, double> pid;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;

const int INF = 0x3f3f3f3f;
const double eps = 1E-6;
const int LEN = 101;
int n, dp[LEN][LEN], num[LEN], sum[LEN];

int main()
{
    freopen("in.txt", "r", stdin);

    int T, kase = 1;
    cin >> T;
    while(T--){
        cin >> n;
        for(int i=1; i<=n; i++)
            cin >> num[i];
        sum[0] = 0;
        for(int i=1; i<=n; i++) sum[i] = sum[i-1] + num[i];
        for(int i=0; i<n; i++){
            for(int j=1; j+i<=n; j++){
                dp[j][i+j] = sum[i+j] - sum[j-1];
                for(int k=j; k<=j+i; k++){
                    if(k!=j)
                        dp[j][i+j] = max(dp[j][i+j], sum[k-1] - sum[j-1] - dp[k][i+j]);
                    if(k!=i+j)
                        dp[j][i+j] = max(dp[j][i+j], sum[i+j] - sum[k] - dp[j][k]);
                }
            }
        }
        cout << "Case " << kase++ << ": " << dp[1][n] << endl;
    }
    return 0;
}