UVA208Firetruck 消防车（图的路径搜索）

题意：输入一个n(n20)个节点的无向图以及某个结点k，按照字典序从小到大顺序输出从结点1到结点k的所有路径，要求经过结点不能重复经过。

分析：终于做到水题了，此题唯一要点就是先判断是否能到达，用dfs。

#include<cstdio>
#include<cstring>
#include<cctype>
#include<queue>
#include<iostream>
#include<vector>
#include<list>
#include<set>
using namespace std;
const int maxn = 20+5;
int G[maxn][maxn], vis[maxn];
int k,m;
int allroute;
int routemap[400][maxn];
void solve(int cur,int dep) {
	if (cur == k) {
		for (int i = 0; i < dep-1; i++) {
			cout << routemap[allroute][i] << " ";
		}
		cout << routemap[allroute][dep - 1];
		cout << endl;
		allroute++;
		memcpy(routemap[allroute], routemap[allroute - 1], sizeof(routemap[allroute - 1]));
		return;
	}
	for (int i = 1; i <= m; i++) {
		if (G[cur][i]&&!vis[i]) {
			vis[i] = 1;
			routemap[allroute][dep] = i;
			solve(i, dep + 1);
			vis[i] = 0;
		}
	}
}
bool isarrive(int cur) {
	if (cur == k) {
		return true;
	}
	for (int i = 1; i <= m; i++) {
		if (G[cur][i] && !vis[i]) {
			vis[i] = 1;
			if (isarrive(i))return true;
		}
	}
	return false;
}
int main() {
	int a, b; int kase = 0;
	while (cin >> k && k) {
		++kase;
		memset(G, 0, sizeof(G));
		memset(vis, 0, sizeof(vis));
		m = 0; allroute = 0;
		while (cin >> a >> b && a&&b) {
			G[a][b] = 1;
			G[b][a] = 1;
			m = max(max(a, b), m);
		}
		cout << "CASE " << kase << ":" << endl;
		if (isarrive(1)) {
			memset(vis, 0, sizeof(vis));
			vis[1] = 1;
			routemap[allroute][0] = 1;
			solve(1, 1);
			cout << "There are "<<allroute<< " routes from the firestation to streetcorner " << k << "." << endl;
		}
		else {
			//不可达
			cout << "There are 0 routes from the firestation to streetcorner " << k << "." << endl;
		}
	}
	//system("pause");
	return 0;
}