LightOJ 1037 - Agent 47 （状压dp）

题意:

给出n个目标,n<=15,再给出一个n∗n的矩阵,开始他对任何一个目标的伤害是1
等他杀死一个目标后,可以使用那个目标的武器来杀死其他目标,问杀死所有目标,最少需要射击多少次

分析:

直接状压dp[s]:=当前剩余敌人状态为s还需要射击多少次
直接n2暴力转移就好了,总复杂度O(n2∗2n),case不多可以接受

代码:

////  Created by TaoSama on 2015-11-13//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, a[15], dp[1 << 15];char b[15][15];int dfs(int s) {    int& ret = dp[s];    if(!s) return 0;    if(~ret) return ret;    ret = INF;    for(int i = 0; i < n; ++i) {        if(s >> i & 1) {            int maxv = 1;            for(int j = 0; j < n; ++j)                maxv = max(maxv, s >> j & 1 ? 0 : b[j][i] - '0');            ret = min(ret, dfs(s ^ (1 << i)) + (a[i] + maxv - 1) / maxv);        }    }    return ret;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        scanf("%d", &n);        for(int i = 0; i < n; ++i) scanf("%d", a + i);        for(int i = 0; i < n; ++i) scanf("%s", b[i]);        memset(dp, -1, sizeof dp);        printf("Case %d: %d\n", ++kase, dfs((1 << n) - 1));    }    return 0;}