POJ Problem 2631 Roads in the North【树的直径】

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原文链接：http://www.cnblogs.com/cniwoq/p/6770873.html

本文介绍了一个算法问题——寻找地图上两个最远村落之间的道路距离。通过构建图结构并使用两次广度优先搜索（BFS），首先找到离起点最远的村落，再以此村落为起点再次进行搜索，从而找出最远距离。

Roads in the North

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2538		Accepted: 1248

Description

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

Sample Output

Source

The UofA Local 1999.10.16

#include <cstring>
#include <cstdio>
#include <queue>
#define MAXN 30010
using namespace std;
struct node{
    int from, to, val, next;
} edge[MAXN];
bool vis[MAXN];
int dist[MAXN], head[MAXN], edgenum, ans, s;
void init() {
    memset(head, -1, sizeof(head));
    edgenum = 0;
}
void addedge(int x, int y, int z) {
    edge[edgenum].from = x;
    edge[edgenum].to = y;
    edge[edgenum].val = z;
    edge[edgenum].next = head[x];
    head[x] = edgenum++;
}
void bfs(int x) {
    queue<int> que; ans = 0;
    memset(vis, false, sizeof(vis));
    memset(dist, 0, sizeof(dist));
    while (!que.empty()) que.pop();
    que.push(x); vis[x] = true;
    while (que.size()) {
        int a = que.front(); que.pop();
        for (int i = head[a]; i != -1; i = edge[i].next) {
            int b = edge[i].to;
            if (!vis[b] && dist[b] < dist[a] + edge[i].val) {
                dist[b] = dist[a] + edge[i].val;
                if(ans < dist[b]) {
                    ans = dist[b]; s = b;
                }
                vis[b] = true; que.push(b);
            }
        }
    }
}
int main() {
    int a, b, c, n, m; init();
    while (scanf("%d%d%d", &a, &b, &c) != EOF) {
        addedge(a, b, c); addedge(b, a, c);
    }
    bfs(1); bfs(s);
    printf("%d\n", ans);
    return 0;
}