Light oj 1037 - Agent 47（状压dp）

1037 - Agent 47

PDF (English)

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Time Limit: 2 second(s)

Memory Limit: 32 MB

Agent 47 is in a dangerous Mission "Black Monster Defeat - 15". It is a secret mission and so 47 has a limited supply of weapons. As a matter of fact he has only one weapon the old weak "KM .45 Tactical (USP)". The mission
sounds simple - "he will encounter at most 15 Targets and he has to kill them all". The main difficulty is the weapon. After huge calculations, he found a way out. That is after defeating a target, he can use target's weapon to kill other targets. So there
must be an order of killing the targets so that the total number of weapon shots is minimized. As a personal programmer of Agent 47 you have to calculate the least number of shots that need to be fired to kill all the targets.



Agent 47
Now you are given a list indicating how much damage each weapon does to each target per shot, and you know how much health each target has. When a target's health is reduced to 0 or less, he is killed.47 start off only with
the KM .45 Tactical (USP), which does damage 1 per shot to any target. The list is represented as a 2D matrix with the ith element containing N single digit numbers ('0'-'9'),
denoting the damage done to targets 0, 1, 2, ..., N-1 by the weapon obtained from target i, and the health is represented as a series of N integers, with the ith element representing
the amount of health that target has.

Given the list representing all the weapon damages, and the health each target has, you should find the least number of shots he needs to fire to kill all of the targets.

Input

Input starts with an integer T (≤ 40), denoting the number of test cases.

Each case begins with a blank line and an integer N (1 ≤ N ≤ 15). The next line contains N space separated integers between 1 and 106 denoting the health of the targets 0,
1, 2, ..., N-1. Each of the next Nlines contains N digits. The jth digit of the ith line denotes the damage done to target j, if you use the
weapon of target i in each shot.

Output

For each case of input you have to print the case number and the least number of shots that need to be fired to kill all of the targets.

Sample Input	Output for Sample Input
2 3 10 10 10 010 100 111 3 3 5 7 030 500 007	Case 1: 30 Case 2: 12

PROBLEM SETTER: MUHAMMAD RIFAYAT SAMEE
SPECIAL THANKS: JANE ALAM JAN

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-8

typedef long long ll;

using namespace std;
#define INF 0x3f3f3f3f
#define N 16

int dp[1<<N];
int a

;
int n;
int va
;

int get(int cur,int j)
{
    int s=1;
    for(int i=0;i<n;i++)
        if(cur&(1<<i))
          s=max(s,a[i][j]);
    return s;
}

void DP()
{
     int i,j;
     int len=1<<n;
     memset(dp,INF,sizeof(dp));
     dp[0]=0;
     for(int i=0;i<len;i++)
         for(j=0;j<n;j++)
         {
             if(i&(1<<j)) continue;
             int t=get(i,j);
             int ti=(va[j]+t-1)/t;
             dp[i|(1<<j)]=min(dp[i|(1<<j)],dp[i]+ti);
         }
}

int main()
{
    int i,j,t,ca=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&va[i]);
        char c
;
        for(i=0;i<n;i++)
        {
            scanf("%s",c);
            for(int j=0;j<n;j++)
                a[i][j]=c[j]-'0';
        }
        DP();
        printf("Case %d: %d\n",++ca,dp[(1<<n)-1]);
    }
    return 0;
}