UVA 208Firetruck

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题目大意：给定节点树不超过20的图，图的节点已数字标识，再给定某个节点的序号，求从节点1到节点k的所有路径，以字典序排序。

这一题的关键在于需要先判断是否能够从节点1到达节点k，也就是使用并查集来判断，同时也需要去除那些不能够到达k的点，这就是剪枝，否则这题会超时。

////  main.cpp//  uva 208 - Firetruck////  Created by XD on 15/7/30.//  Copyright (c) 2015年 XD. All rights reserved.//#include <iostream>#include <string>#include <queue>#include <stack>#include <stdio.h>#include <stdlib.h>#include <math.h>#include<vector>#include <string.h>#include <string>#include <algorithm>#include <set>#include <map>#include <cstdio>using namespace std ;int flag[21] ;int g[21][21] ;int n , k  ;int p[21]  ;int r[21] ;int num ;void init(){    memset(flag, 0, sizeof(flag)) ;          memset(g, 0, sizeof(g)) ;    for (int i = 0; i < 21; i++) {        p[i] = i ;        r[i] = 0 ;    }}int find(int x){    if (p[x]!=x) {        p[x] = find(p[x]) ;    }    return p[x] ;}void link(int x ,int y ){    if (r[x] < r[y]) {        p[x] =y ;    }    else{        p[y] = x ;        if (r[x] == r[y]) {            r[x]+= 1 ;        }    }}void UNION_SET(int x , int y ){    link(find(x), find(y)) ;}int max(int i , int  j ){    return i > j ? i : j ;}int path[20] ;void dfs(int u,int d ){    if (u == k ) {        printf("1" ) ;        int  i = 2 ;        num++ ;        for (i = 1; i < d; i++) {            printf(" %d" ,path[i]) ;        }        printf("\n")   ;        return ;    }    for (int i = 1; i <= n ; i++) {        if (!flag[i] && g[u][i]) {            flag[i]=1 ;            path[d] = i ;            dfs(i , d+1) ;            flag[i] = 0 ;        }    }    }int main() {    int casenum = 0  ;        while (scanf("%d" ,&k)==1) {        num = 0 ;        n = 0  ;        int v ,u  ;        init() ;        printf("CASE %d:\n" , ++casenum) ;        while (scanf("%d%d" ,&v , &u)==2&&u+v!= 0 ) {            n = max(n,max(u,v)) ;            g[u][v] = g[v][u] =1 ;            if (find(u)!= find(v)) {                UNION_SET(u, v) ;            }        }        for (int i = 1; i <=n;i++) {            find(i) ;        }        for (int i = 1; i <=n; i++) {            if (p[k] != p[i]) {                flag[i] = 1 ;            }        }        if (flag[1]) {            printf("There are 0 routes from the firestation to streetcorner %d.\n" , k) ;        }        else{            flag[1] = 1 ;            path[0] = 1 ;            dfs(1,1) ;            printf("There are %d routes from the firestation to streetcorner %d.\n" ,num, k) ;        }    }    return 0;}

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