題干：

Building and maintaining roads among communities in the far North is an expensive business. With this in mind, the roads are build such that there is only one route from a village to a village that does not pass through some other village twice.?
Given is an area in the far North comprising a number of villages and roads among them such that any village can be reached by road from any other village. Your job is to find the road distance between the two most remote villages in the area.?

The area has up to 10,000 villages connected by road segments. The villages are numbered from 1.?

Input

Input to the problem is a sequence of lines, each containing three positive integers: the number of a village, the number of a different village, and the length of the road segment connecting the villages in kilometers. All road segments are two-way.

Output

You are to output a single integer: the road distance between the two most remote villages in the area.

Sample Input

5 1 6 1 4 5 6 3 9 2 6 8 6 1 7

Sample Output

22

解題報告：

? ?這題有好多地方是容易錯的，bfs中的maxx和retp不能不賦值，不然就會wa！想想也是，極限情況，只有一個點，或者是空圖，總之可以構造出一個讓他根本進不去那些循環(huán)的，或者進不去那個while的，所以需要初始化！這也是為什么w=maxx那句不能放到注釋的那里。

AC代碼：

#include <queue> #include <cstdio> #include <cstring> #include <iostream> using namespace std; const int MAX = 2e5 + 5 ; const int INF = 0x3f3f3f3f; struct Node {int to;int w;int ne; } e[MAX]; struct point {int pos,c;point(){}//沒有此構造函數不能寫 node t 這樣point(int pos,int c):pos(pos),c(c){}//可以寫node(pos,cost)這樣}; int head[MAX]; int cnt = 0 ; bool vis[MAX]; void init() {cnt = 0;memset(head,-1,sizeof(head)); } void add(int u,int v,int w) {e[cnt].to = v;e[cnt].w = w;e[cnt].ne = head[u];head[u] = cnt;cnt++; } int bfs(int x,int &w) {queue <point> q;int maxx = 0;int retp = x ;//返回的點坐標 memset(vis,0,sizeof(vis) );q.push(point(x,0));vis[x] = 1;point now;while(q.size() ) {point cur = q.front();q.pop();for(int i = head[cur.pos]; i!=-1; i=e[i].ne) {if(vis[e[i].to]) continue;vis[e[i].to] = 1;now.pos = e[i].to;now.c = cur.c + e[i].w;if(now.c>maxx) {maxx = now.c;retp = now.pos;}q.push(now);}//w = maxx;}w = maxx;return retp; } int main() {init();int u,v,w;while(~scanf("%d%d%d",&u,&v,&w) ) {add(u,v,w);add(v,u,w);}int ans1 = 0,ans2 = 0;u = bfs(1,ans1);v = bfs(u,ans2);printf("%d\n",ans2);return 0 ; }

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