Codeforces Round #222 (Div. 1) 博弈 + dp

一般这种要倒着来。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg

using namespace std;

const int N = 1e5 + 7;
const int M = 1e7 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;

int n, m, cur, x, up, a[N], f[20][1 << 20], op[20], cnt[1 << 20];
char t[20][3];

int dp(int x, int s) {
    if(s == up - 1) return 0;
    if(f[x][s] != inf) return f[x][s];
    f[x][s] = op[x] == 1 ? -inf : inf;
    for(int j = 0; j < n; j++) {
        if((s >> j) & 1) continue;
        int add = t[x][0] == 'p' ? a[j] : 0;
        if(op[x] == 1) f[x][s] = max(f[x][s], dp(x + 1, s | (1 << j)) + add);
        else f[x][s] = min(f[x][s], dp(x + 1, s | (1 << j)) - add);
    }
    return f[x][s];
}

int main() {
    memset(f, inf, sizeof(f));
    scanf("%d", &n);
    for(int i = 0; i < n; i++) scanf("%d", &a[i]);
    sort(a, a + n); reverse(a, a + n);
    scanf("%d", &m); n = min(n, m); up = 1 << n;

    for(int i = 0; i < m; i++)
        scanf("%s%d", t[i], &op[i]);

    printf("%d\n", dp(0, 0));
    return 0;
}


/*
*/

 

posted @ 2018-08-23 15:12  NotNight  阅读(179)  评论(0编辑  收藏  举报