Uva - 10891 - Game of Sum-优快云博客

探讨了两人游戏中的策略，玩家A与B从一列数中取数，直到所有数被取完，分析了如何最大化自己的分数。通过定义状态转移方程，实现了算法解决此类问题。

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题意：A与B玩游戏，从一列数，n个（0 < n <= 100）中，每次可以从左端或者右端连续的取几个数，直到所有数被取完，问最后A取的数的和减去B取的数的和是多少。

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=20&problem=1832

——>>设d[i][j]为下标从i到j的序列中先手最高得分。

状态转移方程：

d[i][j] = sum[i][j] - min(d[i+1][j], d[i+2][j], ..., d[j][j], d[i][j-1], d[i][j-2], ..., d[i][i], 0);

设l[i][j] = min(d[i][j], d[i+1][j], d[i+2][j], ..., d[j][j]);

则l[i][j] = min(d[i][j], l[i+1][j]);

r[i][j] = min(d[i][j], d[i][j-1], d[i][j-2], ..., d[i][i]);

则r[i][j] = min(d[i][j], r[i][j-1]);

则d[i][j] = sum[i][j] - min(l[i+1][j], r[i][j-1], 0);

#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn = 100 + 10;

int main()
{
    int n, a[maxn], sum[maxn], l[maxn][maxn], r[maxn][maxn], d[maxn][maxn], i, j;
    while(scanf("%d", &n) == 1 && n)
    {
        sum[0] = 0;
        for(i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
            sum[i] = sum[i-1] + a[i];
        }
        for(i = 1; i <= n; i++) d[i][i] = l[i][i] = r[i][i] = a[i];     //边界
        for(int L = 1; L < n; L++)      //递推
            for(i = 1; i+L <= n; i++)
            {
                j = i + L;
                int m = 0;
                m = min(m, l[i+1][j]);
                m = min(m, r[i][j-1]);
                d[i][j] = sum[j] - sum[i-1] - m;
                l[i][j] = min(d[i][j], l[i+1][j]);
                r[i][j] = min(d[i][j], r[i][j-1]);
            }
        printf("%d\n", 2*d[1][n] - sum[n]);
    }
    return 0;
}