uva_10891 - Game of Sum( 博弈区间DP )
2012-12-04 15:25
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这题是一个区间DP,其实题目已经很明显了,只能从两端取+重复子问题,这样挺容易想到区间DP的。 状态: dp[i][j][A] 表示在区间【i,j]上A先取获得的最大差值, 同理dp[i][j]【B】 状态转移: dp[i][j][A/B] = max(sum[i, k]+dp[k+1][j][A/B], sum[k', j]+dp[i, k'-1][A/B]) #include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define A 0 #define B 1 #define PLAYER 2 #define MAXN 101 #define INF 0x3f3f3f3f int val[MAXN], dp[MAXN][MAXN][PLAYER], sum[MAXN][MAXN]; int dfs(const int &l, const int &r, const int &flag) { if( l >= r ) { return (l == r)? val[l] : 0; } if( -1 != dp[l][r][flag] ) { return dp[l][r][flag]; } int rst; rst = dp[l][r][flag] = -INF; for(int i = l; i <= r; i ++) { if( rst < sum[l][i]+sum[i+1][r]-2*dfs(i+1, r, !flag) ) { dp[l][r][flag] = max(dp[l][r][flag], sum[l][i]+sum[i+1][r]-dfs(i+1, r, !flag)); } } for(int i = l+1; i <= r; i ++) { if( rst < sum[i][r]+sum[l][i-1]-2*dfs(l, i-1, !flag) ) { dp[l][r][flag] = max(dp[l][r][flag], sum[i][r]+sum[l][i-1]-dfs(l, i-1, !flag)); } } return dp[l][r][flag]; } int main(int argc, char const *argv[]) { #ifndef ONLINE_JUDGE freopen("test.in", "r", stdin); #endif int n; while( scanf("%d", &n) && n ) { for(int i = 0; i < n; i ++) { scanf("%d", &val[i]); } memset(sum, 0, sizeof(sum)); memset(dp, -1, sizeof(dp)); for(int i = 0, tmp = 0; i < n; i ++, tmp = 0) { for(int j = i; j < n; j ++) { tmp += val[j]; sum[i][j] = tmp; } } printf("%d\n", 2*dfs(0, n-1, A)-sum[0][n-1]); } return 0; }
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