有两个双端队列每次每人可以从任意一堆的头或尾取值,两人均采取最优策略,问先手的最后得分
很简单的博弈
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int inf = 1000000000;
int T, N, A[25], B[25], f[25][25][25][25];
int DP(int a, int b, int c, int d) {
int turn = N - (b - a + 1) + N - (d - c + 1);
if(turn == (N << 1)) return 0;
int & res = f[a][b][c][d];
if(res != -1) return res;
if(turn & 1) {
res = inf;
if(a <= b) res = min(res, DP(a + 1, b, c, d));
if(a <= b) res = min(res, DP(a, b - 1, c, d));
if(c <= d) res = min(res, DP(a, b, c + 1, d));
if(c <= d) res = min(res, DP(a, b, c, d - 1));
}
else {
res = 0;
if(a <= b) res = max(res, DP(a + 1, b, c, d) + A[a]);
if(a <= b) res = max(res, DP(a, b - 1, c, d) + A[b]);
if(c <= d) res = max(res, DP(a, b, c + 1, d) + B[c]);
if(c <= d) res = max(res, DP(a, b, c, d - 1) + B[d]);
}
return res;
}
int main() {
scanf("%d", &T);
for(int kase = 1; kase <= T; kase++) {
scanf("%d", &N);
for(int i = 1; i <= N; i++) {
scanf("%d", &A[i]);
}
for(int i = 1; i <= N; i++) {
scanf("%d", &B[i]);
}
memset(f, -1, sizeof(f));
printf("%d
", DP(1, N, 1, N));
}
return 0;
}