uva10160(dfs+状态压缩）

Problem D: Servicing stations

A company offers personal computers for sale in N towns (3 <= N <= 35). The towns are denoted by 1, 2, ..., N. There are direct routes connecting M pairs from among these towns. The company decides to build servicing
stations in several towns, so that for any town X, there would be a station located either in X or in some immediately neighbouring town of X.
Write a program for finding out the minumum number of stations, which the company has to build, so that the above condition holds.
Input

The input consists of more than one description of town (but totally, less than ten descriptions). Every description starts with number N of towns and number M of pairs of towns directly connected each other.
The integers N and M are separated by a space. Every one of the next M rows contains a pair of connected towns, one pair per row. The pair consists of two integers for town's numbers, separated by a space. The input ends with N = 0 and M = 0.

Output

For every town in the input write a line containing the obtained minimum.

An example:

Input:
8 12

1 2

1 6

1 8

2 3

2 6

3 4

3 5

4 5

4 7

5 6

6 7

6 8

0 0
Output:
2
参考了点击打开链接,多添加了|结果wa了好多次

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define ll long long

int n, m;
ll a[40], b[40];

bool dfs(ll state, int cur, int next, int num){
if(state== ((ll)1<<n)-1){
return true;
}
if(cur == num){
return false;
}
for(int i = next; i <= n; i++){
if((state|a[i])==state){
continue;
}
if((state|b[i])!=((ll)1<<n)-1){
return false;
}
if(dfs(state|a[i], cur+1, i+1, num)){
return true;
}
}
return false;
}

int main(){
//	freopen("in.txt", "r", stdin);
while(scanf("%d%d", &n, &m),n&&m){
//		memset(a, 0, sizeof(a));
for(int i = 1; i <=n; i++){
a[i] = ((ll)1<<(i-1));
}
int tmp, tmp2;
for(int i = 1; i <= m; i++){
scanf("%d%d", &tmp, &tmp2);
a[tmp] = a[tmp] |((ll)1<<(tmp2-1));
a[tmp2] = a[tmp2] |((ll)1<<(tmp-1));
}
b
= a
;
for(int i = n-1; i>0; i--){
b[i] = a[i]|b[i+1];
}
for(int i = 1; i <= n; i++){
if(dfs(0,0,1,i)){
printf("%d\n", i);
break;
}
}
}
return 0;
}