lightoj 1037 - Agent 47 【状压dp】

题目链接：lightoj 1037 - Agent 47

1037 - Agent 47
PDF (English) Statistics Forum
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.

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

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

题意：给定要杀掉的n个人的H值，每次你可以打掉1点H值。又给出一个n*n的矩阵，Map[i][j]表示第i个人的武器可以一次打掉第j个人Map[i][j]点H值。问你最少需要多少次可以杀掉这n个人。

思路：n <= 15显然状压。我们每次枚举状态s下最后一个被杀死的人，dp[s]=min(dp[s−(1<<i)]+val[s−(1<<i)][i],dp[s])这样时间复杂度O(n2∗2n)。val[s][i]表示s状态下杀死第i个人的最少次数。
AC代码：

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <iostream>#define CLR(a, b) memset(a, (b), sizeof(a))using namespace std;const int INF = 0x3f3f3f3f;int Map[20][20], val[1<<15][20];int dp[1<<15], H[20];int Solve(int n, int m) {    if(m == 0) return INF;    if(n % m == 0) return n / m;    else return n / m + 1;}int main(){    int t, kcase = 1;    scanf("%d", &t);    while(t--) {        int n; scanf("%d", &n);        for(int i = 0; i < n; i++ ) {            scanf("%d", &H[i]);        }        for(int i = 0; i < n; i++ ) {            char str[20]; scanf("%s", str);            for(int j = 0; j < n; j++) {                Map[i][j] = str[j] - '0';            }        }        for(int i = 0; i < (1<<n); i++) {            for(int j = 0; j < n; j++) {                val[i][j] = H[j];            }        }        for(int i = 1; i < (1<<n); i++) {            for(int j = 0; j < n; j++) {                if(i & (1<<j)) {                    int s = i ^ (1<<j);                    for(int k = 0; k < n; k++) {                        if(s & (1<<k)) {                            val[s][j] = min(val[s][j], Solve(H[j], Map[k][j]));                        }                    }                }            }        }        dp[0] = 0;        for(int i = 1; i < (1<<n); i++) {            dp[i] = INF;            for(int j = 0; j < n; j++) {                if(i & (1<<j)) {                    int s = i ^ (1<<j);                    dp[i] = min(dp[i], dp[s] + val[s][j]);                }            }        }        printf("Case %d: %d\n", kcase++, dp[(1<<n)-1]);    }    return 0;}