uva-10160-枚举

若当前搜索到的城市n前面1～n-1编号的城市中有没有通电的，则永远也无法输送电力给那个城市,因为在剪枝时附加了和此结点连接的最大结点小于当前的结点

这段

for(int i = 1; i < cur; i++)
if(!vis[i] && map[i][0] < cur)
return;

#include<iostream>
#include <new>
#include<stdio.h>
#include<memory.h>
using namespace std;
int n, m;
int map[100][100];
int index2[100];
int vis[100];
int res = 0;
void sort(int a[], int length)
{
    for(int i = 0; i < length; i++)
        for(int j = 1; j < length - i; j++)
            if(a[j] > a[j - 1])
            {
                int k = a[j - 1];
                a[j - 1] = a[j];
                a[j] = k;
            }
}
void dfs(int cur, int sum, int total)
{
    if(sum >= res)
        return;
    if(total == n)
    {
        res = sum;
        return;
    }
    for(int i = 1; i < cur; i++)
        if(!vis[i] && map[i][0] < cur)
            return;
    //不选
    dfs(cur + 1, sum, total);
    int k = 0;
    int vis2[100] = { 0 };
    for(int i = 0; i < index2[cur]; i++)
    {
        if(!vis[map[cur][i]])
        {
            vis[map[cur][i]] = 1;
            //选,这次覆盖到点
            vis2[k++] = map[cur][i];
        }
    }
    if(!k)
        //没有新的点覆盖
        return;
    dfs(cur + 1, sum + 1, total + k);
    for(int i = 0; i < k; i++)
        vis[vis2[i]] = 0;
}
int main()
{
    freopen("d:\1.txt", "r", stdin);
    while (cin >> n >> m && n != 0 && m != 0)
    {
        memset(map, -1, sizeof(map));
        memset(vis, 0, sizeof(vis));
        memset(index2, 0, sizeof(index2));
        int s, e;
        for(int i = 0; i < m; i++)
        {
            cin >> s >> e;
            map[s][index2[s]++] = e;
            map[e][index2[e]++] = s;
        }
        for(int i = 1; i <= n; i++)
        {
            map[i][index2[i]++] = i;
            sort(map[i], index2[i]);
        }
        res = 0x7fffffff;
        dfs(1, 0, 0);
        cout << res << endl;
    }
    return 0;
}