uva 208(回溯)

题意：给出一个目标节点，和各个节点的连通情况，然后寻找从1到这个目标节点的所有可能路径。

题解：直接回溯会TLE。。。所以需要剪枝，可以将所有能到达目标节点的节点放在一起再排个序，保证了图的连通性，这样就不会在某个点死循环了。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 22;

int g[N][N], ans, res[N], vis[N], n, num, node[N], tar;

void init() {
	memset(g, 0, sizeof(g));
	memset(res, 0, sizeof(res));
	memset(vis, 0, sizeof(vis));
	ans = num = n = 0;
}

void search(int cur) {
    node[num++] = cur;
    vis[cur] = 1;
    for (int i = 1; i <= n; i++)
        if (g[cur][i] && !vis[i]) {
            search(i);
        }
}

void dfs(int cur, int count) {
    if (cur == tar) {
        printf("%d", res[0]);
		for (int i = 1; i < count; i++)
			printf(" %d", res[i]);
		printf("\n");
		ans++;
		return;
    }
	for (int i = 0; i < num; i++)
		if (g[cur][node[i]] && !vis[node[i]]) {
			vis[node[i]] = 1;
			res[count] = node[i];
			dfs(node[i], count + 1);
			vis[node[i]] = 0;
		}
}

int main() {
	int a, b;
	int t = 1;
	while (scanf("%d", &tar) != EOF) {
		init();
		while (scanf("%d%d", &a, &b) && a) {
			if (a > n)
				n = a;
			if (b > n)
				n = b;
			g[a][b] = g[b][a] = 1;
		}
		printf("CASE %d:\n", t++);
		search(tar);
		sort(node, node + num);
		memset(vis, 0, sizeof(vis));
		vis[1] = 1;
		res[0] = 1;
		dfs(1, 1);
		printf("There are %d routes from the firestation to streetcorner %d.\n", ans, tar);
	}
	return 0;
}