poj 1041 John's trip (边最小字典序欧拉路径 Fleury)

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题目链接: poj 1041

题目大意: 给出无向图，每条边有唯一的序号

是否存在欧拉回路，若存在输出边序号最小字典序的路径

解题思路: 无向欧拉回路的判断方法，若存在奇数度点，则不存在欧拉回路

依据静态邻接链表的特性，从序号大到小建立边

因为这样总能保证序号最小的边首先访问到

PS:欧拉路径的问题要记得判断图是否联通

代码:

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <algorithm>using namespace std;#define MAX2010#define INF 0x3f3f3f3fint S,E,n,Index,pre[MAX],visit[MAX],Du[MAX],kk,answer[MAX];struct snode{int w,pd,to,next;}edge[MAX<<3];struct node{int a,b,c;}ans[MAX<<3];bool cmp(struct node a,struct node b){return (a.c<b.c)?1:0;}void Add_edge(int a,int b,int c){edge[Index].to=b;edge[Index].w=c;edge[Index].next=pre[a];pre[a]=Index++;}void Fleury(int start){int i,v;for(i=pre[start];i!=-1;i=edge[i].next){v=edge[i].to;if(!visit[edge[i].w]){visit[edge[i].w]=1;Fleury(v);answer[++kk]=edge[i].w;  //记录边的访问顺序}}}int main(){int i,k,a1,a2,a3,pd,start;while(scanf("%d%d",&a1,&a2)!=EOF){k=1,Index=kk=pd=0;if(a1==0&&a2==0)break;memset(visit,0,sizeof(visit));memset(ans,0,sizeof(ans));memset(pre,-1,sizeof(pre));memset(Du,0,sizeof(Du));scanf("%d",&a3);ans[k].a=a1,ans[k].b=a2,ans[k].c=a3;Du[a1]++,Du[a2]++;        if(a1>a2)            S=a2,E=a1;        else            S=a1,E=a2;while(1){scanf("%d%d",&a1,&a2);if(a1==0&&a2==0)break;scanf("%d",&a3);if(a1<S) S=a1;if(a2<S) S=a2;if(a1>E) E=a1;if(a2>E) E=a2;Du[a1]++,Du[a2]++;ans[++k].a=a1,ans[k].b=a2,ans[k].c=a3;}start=S;for(i=S;i<=E;i++){if(Du[i]==0)continue;if(Du[i]%2==1)   //有奇数度点则不存在欧拉回路{pd=1;break;}}if(pd==1){printf("Round trip does not exist.\n");continue;}        sort(ans+1,ans+1+k,cmp);for(i=k;i>=1;i--){Add_edge(ans[i].a,ans[i].b,ans[i].c);Add_edge(ans[i].b,ans[i].a,ans[i].c);}Fleury(start);for(i=kk;i>=1;i--)printf("%d%c",answer[i],(i==1)?'\n':' ');}return 0;}

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