UVA 10891 Game of Sum (区间DP)

题意：有n个数，两个玩家A、B，每个玩家在他那一轮可以从左端或右端取连续的一段数，把数的和加到自己的分数上，假设每个玩家都采取尽可能的让自己的分数更高，那么A玩家最后和B玩家分数的差为多少（A玩家先手）

思路：

设dp[i][j][0]为自己先手在区间i到j内最后获得的最多的分数

dp[i][j][1] 为对方先手在区间i到j内最后获得的最多的分数

那么状态转移方程：

dp[i][j][0] = max(dp[i][j][0],max(sum[k] - sum[i-1] + dp[k+1][j][1],sum[j] - sum[k-1] + dp[i][k-1][1]));

dp[i][j][1] = min(dp[i][j][1],min(dp[k+1][j][0],dp[i][k-1][0]));

最后结果为dp[0][n-1][0] - dp[0][n-1][1].

我的代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;
const int maxn = 105;
const int INF = 0x3f3f3f3f;

int n,a[maxn],sum[maxn];
int dp[maxn][maxn][2];

int main(){
while(~scanf("%d",&n)){
if(n == 0) break;
memset(dp,0,sizeof(dp));
memset(sum,0,sizeof(sum));
for(int i = 0 ; i < n ; i++){
scanf("%d",&a[i]);
for(int j = 0 ; j <= i ; j++)
sum[i] += a[j];
}
for(int i = n - 1 ;i >= 0 ; i--){
for(int j = i ; j < n ; j++){
dp[i][j][0] = -INF ; dp[i][j][1] = INF;
if(i == j){
dp[i][j][0] = a[i];
dp[i][j][1] = 0;
continue;
}
for(int k = i ; k <= j ; k++){
if(i == 0 && k == 0){
dp[i][j][0] = max(dp[i][j][0],max(sum[k] + dp[k+1][j][1],sum[j]));
dp[i][j][1] = min(dp[i][j][1],min(dp[k+1][j][0],0));
}else if(i == 0){
dp[i][j][0] = max(dp[i][j][0],max(sum[k] + dp[k+1][j][1],sum[j] - sum[k-1] + dp[i][k-1][1]));
dp[i][j][1] = min(dp[i][j][1],min(dp[k+1][j][0],dp[i][k-1][0]));

}else{
dp[i][j][0] = max(dp[i][j][0],max(sum[k] - sum[i-1] + dp[k+1][j][1],sum[j] - sum[k-1] + dp[i][k-1][1]));
dp[i][j][1] = min(dp[i][j][1],min(dp[k+1][j][0],dp[i][k-1][0]));
}
}
}
}

printf("%d\n",dp[0][n-1][0] - dp[0][n-1][1]);
}
return 0;
}