1037 - Agent 47(状压DP)
2016-03-07 22:03
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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.
182011-agent47_large
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 N lines 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
杀手去杀n个人,(n <= 15),杀完这个人可以拿被杀死这个人的武器,每个目标的武器对彼此也有一定的伤害,杀手自带的手枪伤害始终为1,每个人有血量,把血量打完才算打死,问最少开几枪能把目标清除完。
状压啊,一开始把0放队列里,然后每次取出队首,这个数0代表没死,1代表死了,枚举0,然后用杀死的人和自身武器最大伤害打目标,然后不断转移= =
182011-agent47_large
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 N lines 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
杀手去杀n个人,(n <= 15),杀完这个人可以拿被杀死这个人的武器,每个目标的武器对彼此也有一定的伤害,杀手自带的手枪伤害始终为1,每个人有血量,把血量打完才算打死,问最少开几枪能把目标清除完。
状压啊,一开始把0放队列里,然后每次取出队首,这个数0代表没死,1代表死了,枚举0,然后用杀死的人和自身武器最大伤害打目标,然后不断转移= =
#include<cstdio> #include<cstring> #include<iostream> #include<queue> #include<vector> #include<algorithm> #include<string> #include<cmath> #include<set> #include<map> #include<vector> #include<stack> #include<utility> #include<sstream> using namespace std; typedef long long ll; const int inf = 0x3f3f3f3f; const int maxn = (1 << 16) + 10; int t,n; int a[20],b[20][20],dp[maxn]; int main() { #ifdef LOCAL freopen("C:\\Users\\ΡΡ\\Desktop\\in.txt","r",stdin); //freopen("C:\\Users\\ΡΡ\\Desktop\\out.txt","w",stdout); #endif // LOCAL int kase = 1; scanf("%d",&t); while(t--) { scanf("%d",&n); for(int i = 1;i <= n;i++) scanf("%d",a + i); for(int i = 1;i <= n;i++) { char s[20]; scanf("%s",s); for(int j = 0;j < n;j++) b[i][j + 1] = s[j] - '0'; } for(int i = 0;i <= (1 << n);i++) dp[i] = inf; dp[0] = 0; queue<int>p; p.push(0); while(p.size()) { int u = p.front();p.pop(); for(int i = 0;i < n;i++) { if((u >> i)&1)continue; int k = 1; for(int j = 0;j < n;j++) { if((u >> j)&1) k = max(k,b[j + 1][i + 1]); } int now = u | (1 << i); if(dp[now] > dp[u] + (a[i + 1] + k - 1)/k) { dp[now] = dp[u] + (a[i + 1] + k - 1)/k; p.push(now); } } } printf("Case %d: %d\n",kase++,dp[(1 << n) - 1]); } return 0; }
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