uva10891 - Game of Sum(动归)

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状态：dp[i][j]表示从a[i]到a[j]的串中，第一个选取的人所能选出的最大值。

状态转移：dp[i][j] = sum[i....j] - min(dp[i+1][j],dp[i+2][j],dp[i+3][j],dp[i+4][j]....dp[j][j], dp[i][j-1],dp[i][j-1],dp[i][j-3], dp[i][j-4]...dp[i][i], 0);

“0”是代表第一个人把a[i]....a[j]的全取的状态：

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define N 105int n, a[N], dp[N][N], f[N][N], g[N][N], s[N];int main (){    while(scanf("%d",&n),n)    {        s[0] = 0;        memset(dp,0,sizeof(dp));        for(int i = 1; i <= n; i++) {scanf("%d",&a[i]); s[i] = s[i-1]+a[i];}        for(int i = 1; i <= n; i++) dp[i][i] = f[i][i] = g[i][i] = a[i];        for(int l = 1; l < n; l++)        for(int i = 1, j = i+l; j <= n; i++, j++)        {            int m = 0;            m = min(m, f[i+1][j]);            m = min(m, g[i][j-1]);            dp[i][j] = s[j]-s[i-1]-m;            f[i][j] = min(dp[i][j],f[i+1][j]);            g[i][j] = min(dp[i][j],g[i][j-1]);        }        printf("%d\n",2*dp[1][n]-s[n]);    }    return 0;}