UVA - 208 Firetruck

题目大意：给一个点，然后给一系列相连的线段，要求按字典序输出从1到该点所经过的点

解题思路：这个如果直接回溯的话，会出现TLE，因为有做无用功，如果是稠密图的话，就会出错了，先把和终点有联系的点找出来，然后将其放在一个数组中，排序一下，再用该数组进行判断，判断点是否符合要求

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 30
int g[maxn][maxn], res[maxn], vis[maxn], link[maxn], judge[maxn];
int _max,num,_count,number;

void dfs_1(int cur) {
	judge[cur] = 1;
	link[_count++] = cur;
	for(int i = 1; i <= _max; i++)
		if(!judge[i] && g[cur][i])
			dfs_1(i);	
}

void dfs_2(int cur, int n) {
	if(cur == num) {
		for(int i = 0 ; i < n - 1; i++)
			printf("%d ",res[i]);
		printf("%d\n",res[n-1]);	
		number++;
	}
	else 
		for(int i = 1; i < _count; i++) 
			if(vis[link[i]] && g[cur][link[i]]) {
				vis[link[i]] = 0;
			   	res
 = link[i];
				dfs_2(link[i],n+1);
				vis[link[i]] = 1;
			}
}

int main() {
	int mark = 1;
	int n1, n2;

	while(scanf("%d", &num) != EOF) { 

		memset(g,0,sizeof(g));
		memset(judge,0,sizeof(judge));
		memset(link,0,sizeof(link));
		_max = 0; _count = 1; number = 0;

		for(int i = 0; ;i++) {
			scanf("%d%d", &n1,&n2);
			if(!n1 && !n2)
				break;

			g[n1][n2]++;
			g[n2][n1]++;
			if(n1 > _max)
				_max = n2;
			if(n2 > _max)
				_max = n2;
		}

		printf("CASE %d:\n",mark++);	

		for(int i = 1; i <= _max ; i++)
			vis[i] = 1;

		res[0] = 1;
		vis[1] = 0;
		dfs_1(num);
		sort(link,link+_count);
		dfs_2(1,1);
		printf("There are %d routes from the firestation to streetcorner %d.\n",number,num);
	}	
	return 0;
}