Light 1037 - Agent 47 (狀態壓縮dp)

題意

每次打敗一個怪獸可以獲得某個手槍。問最少開幾槍。

思路

dp[S]表示打敗S狀態的怪獸最少多少槍。

對於一個狀態S，我們枚舉有哪些怪獸在裏面，然後一一去除，選出打那隻怪獸最高攻擊力的轉移一下。

代碼

#include <stack>
#include <cstdio>
#include <list>
#include <cassert>
#include <set>
#include <fstream>
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <string>
#include <map>
#include <cmath>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
//using namespace __gnu_pbds;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(p, num) memset(p, num, sizeof(p))
#define PB push_back
#define X first
#define Y second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define FOR(i, a, b) for (int i=(a); (i) < (b); (i)++)
#define FOOR(i, a, b) for (int i = (a); (i)<=(b); (i)++)
#define TRAVERSAL(u, i) for (int i = head[u]; i != -1; i = edge[i].nxt)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 1e5+10;
const int MOD = 1e9+7;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int seed = 131;
int cases = 0;
typedef pair<int, int> pii;

int dp[1<<16], life[20], n;
char atk[20][20];

int dfs(int S)
{
    if (dp[S] != INF) return dp[S];
    for (int i = 0; i < n; i++) if ((S>>i)&1)  //從哪個狀態轉移過來
    {
        int max_atk = 1;
        for (int j = 0; j < n; j++)
        {
            if (j == i || !((S>>j)&1)) continue;
            max_atk = max(max_atk, (int)atk[j][i]);
        }
        dp[S] = min(dp[S], dfs(S^(1<<i))+(life[i]+max_atk-1)/max_atk);
    }
    return dp[S];
}

int main()
{
    //ROP;
    int T;
    scanf("%d", &T);
    while (T--)
    {
        MS(dp, INF);
        dp[0] = 0;
        scanf("%d", &n);
        for (int i = 0; i < n; i++) scanf("%d", &life[i]);
        for (int i = 0; i < n; i++)
        {
            scanf("%s", atk[i]);
            for (int j = 0; j < n; j++) atk[i][j] -= '0';
        }
        printf("Case %d: %d\n", ++cases, dfs((1<<n)-1));
    }
    return 0;
}