UVA10891[Game of Sum] 动态规划

题目链接

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题目大意：两个玩家A和B在玩一个取石子游戏，且每个石子都有它们各自的价值。在游戏中有这样一个规则：每次取一个石子或者连续几个都必从两端取，要么是最左端，要么是最右端，直到取完为止。两个玩家都非常聪明，他们每次都会去最优的结果。给他们N个石子，你能计算出玩家A，B各自的最后结果吗？假设总是玩家A先开局。

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解题报告：

dp[i][j] 表示对方剩下的序列为原序列的i~j子序列。

dp[i][j]=sum[i][j]-min{d[i+1][j],d[i+2][j]…,d[j][j],…,d[i][j-1]+d[i][i],0}

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int N = 100 + 10;int S[N], A[N], d[N][N], vis[N][N], n;int dp(int i, int j){    if( vis[i][j] ) return d[i][j];    vis[i][j]=1;    int m=0;    for ( int k=i+1; k<=j; k++ ) m=min( m, dp(k,j) );    for ( int k=i; k<j; k++ ) m=min( m, dp(i,k) );    d[i][j]=S[j]-S[i-1]-m;    return d[i][j];} int main(){    while( scanf("%d", &n)==1 && n ){        S[0]=0;        for ( int i=1; i<=n; i++ ) scanf("%d", &A[i] ), S[i]=S[i-1]+A[i];        memset(vis,0,sizeof(vis));        dp(1,n);        printf("%d\n", 2*d[1][n]-S[n] );    }    return 0;}

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f[i][j]=min(dp[i[j],f[i+1][j])

g[i][j]=min(dp[i][j],f[i][j-1])

d[i][j]=sum(i,j)-min(f[i+1][j],g[i][j-1],0)

#include <cstdio>#include <iostream>#include <cstring>using namespace std;const int N = 110;int S[N], A[N], d[N][N], f[N][N], g[N][N];int main(){    int n;    while( scanf("%d", &n )==1 && n ){        S[0]=0;        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++ ) d[i][i]=f[i][i]=g[i][i]=A[i];        for ( int L=1; L<n; L++ )            for ( int i=1; i+L<=n; i++ ){                int j=i+L;                int m=0;                m=min(m, f[i+1][j]);                m=min(m, g[i][j-1]);                d[i][j]=S[j]-S[i-1]-m;                f[i][j]=min(d[i][j], f[i+1][j]);                g[i][j]=min(d[i][j], g[i][j-1]);            }         printf("%d\n", 2*d[1][n]-S[n] );    }    return 0;}