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7 1 2 1 2 3 2 3 4 1 4 5 1 3 6 3 6 7 4 0
5
ac代码
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<string>
#include<iostream>
#define INF 0xfffffff
#define min(a,b) (a>b?b:a)
using namespace std;
int n,ans;
struct node
{
int u,v,w,next;
}edge[20200];
int head[10010],cnt,vis[10010],dig[10010];
void add(int u,int v,int w)
{
edge[cnt].u=u;
edge[cnt].v=v;
edge[cnt].w=w;
edge[cnt].next=head[u];
head[u]=cnt++;
}
struct s
{
int x,step;
friend bool operator < (s a,s b)
{
return a.step>b.step;
}
}a,temp;
int jud(struct s a)
{
if(vis[a.x])
return 0;
if(a.step>ans)
return 0;
return 1;
}
int bfs(int x)
{
a.x=x;
a.step=0;
priority_queue<struct s>q;
q.push(a);
memset(vis,0,sizeof(vis));
vis[x]=1;
while(!q.empty())
{
a=q.top();
q.pop();
int u=a.x;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
int w=edge[i].w;
temp.x=v;
temp.step=a.step+w;
if(!jud(temp))
continue;
if(dig[v]==1)
return temp.step;
vis[v]=1;
q.push(temp);
}
}
return ans;
}
int main()
{
while(scanf("%d",&n)!=EOF,n)
{
int i;
cnt=0;
memset(head,-1,sizeof(head));
memset(dig,0,sizeof(dig));
for(i=0;i<n-1;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
dig[u]++;
dig[v]++;
}
ans=INF;
for(i=1;i<=n;i++)
{
if(dig[i]==1)
{
int tep=bfs(i);
ans=min(ans,tep);
}
}
printf("%d\n",ans);
}
}
HDOJ 题目3848 CC On The Tree(BFS)
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原文地址:http://blog.csdn.net/yu_ch_sh/article/details/45396141