B - Agent 47

Description

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.

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 i^th 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 Nintegers, with the i^th 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 10⁶denoting the health of the targets 0, 1, 2, ..., N-1. Each of the next N lines contains N digits. The j^th digit of the i^th 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

2

 

3

10 10 10

010

100

111

 

3

3 5 7

030

500

007

Sample Output

Case 1: 30

Case 2: 12

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int inf = ~0u>>2;int dp[1<<15];int hp[15];int two[16];char mp[20][20];int a[20][20];int main(){    two[0] = 1;for(int i = 1; i < 16; i++) two[i] = two[i-1] * 2;int t,ca=1,n;scanf("%d",&t);while(t--){    scanf("%d",&n);for(int i = 0; i < n; i++) scanf("%d",&hp[i]);for(int i = 0; i < n; i++){scanf("%s",mp[i]);for(int j = 0; j < n; j++){a[i][j] = mp[i][j] - '0';}}fill(dp,dp+two[n],inf);dp[0] = 0;for(int i = 0; i < n ;i++) dp[two[i]] = hp[i]; for(int i = 0; i < two[n]; i++)  //i表示当前杀过人的状态，000101表示第三个人和第一个人被杀 {    for(int j = 0; j < n; j++) if(dp[i]!=inf)if(i&two[j])//j表示现在对第几个人下黑手（死掉的不管他） {            //i&two[j] 如果等于1表示当前状态下第j个人是还活着，下一步找出对他伤害最大的武器     for(int k = 0; k < n; k++) //k表示以前被灭掉的家伙，找出缴获的枪 {if(!(i&two[k]) && a[j][k])   //!(i&two[k]) 等于1表示这个人已经被灭掉，a[j][k]表示他的枪 dp[i|two[k]] = min(dp[i|two[k]],dp[i]+(hp[k]+a[j][k]-1)/a[j][k]);//如果这个人血量5，枪的伤害为3，开两枪 dp[i|two[k]] = min(dp[i|two[k]],dp[i]+hp[k]);}}}printf("Case %d: %d\n",ca++,dp[two[n]-1]);}return 0;}