Hlg 树的直径.cpp

题意：
　　给出一棵树..求出一条路..使得别的点到这条路径的距离最长的最短..

输入：
　　　　n 表示树上有n个点..
　　　　u v w 表示u和v这两个相连的点的边权值为w

思路：
　　3次广搜..
　　第一次从任意一个点找最长的路径~这个路径的终点就是要求的路的终点..
　　第二次从终点开始找最长的路径..这时候找到的路径就是最后的路了..
　　然后从这个队列往外遍历相连的路..
　　这时候找出来的最长的距离就是要求的答案..

Tips：
　　因为加边的时候是双向边..
　　所以边的数组应该开两倍大..
　　pre 数组要注意~~~~

Code：

#include <iostream>
#include <cstring>
#include <stdio.h>
using namespace std;
#define clr(x) memset(x, 0, sizeof(x))

const int MAXN = 100010;

struct Edge
{
    int to;
    int next;
    int w;
}edge[MAXN*2];
int tot;
int head[MAXN];

void add(int s, int u, int w) {
    edge[tot].to = u;
    edge[tot].next = head[s];
    edge[tot].w = w;
    head[s] = tot++;

    edge[tot].to = s;
    edge[tot].next = head[u];
    edge[tot].w = w;
    head[u] = tot++;
}

struct Node
{
    int num;
    int dis;
}que[MAXN];

bool vis[MAXN];
int rd[MAXN];
int bfs()
{
    Node tmp, tmp1;
    int pre[MAXN];
    int tmps, ma = 0, s, e, ans = 0;
    clr(pre);
    int front = 0, rear = 0;
    clr(vis);
    tmp.num = 1;
    tmp.dis = 0;
    que[rear++] = tmp;
    vis[1] = true;
    while(front < rear) {
        tmp = que[front++];
        tmps = tmp.num;
        for(int i = head[tmps]; i; i = edge[i].next)
        if(!vis[edge[i].to]) {
            tmp1.num = edge[i].to;
            tmp1.dis = edge[i].w + tmp.dis;
            que[rear++] = tmp1;
            vis[edge[i].to] = true;
            if(ma < tmp1.dis) {
                ma = tmp1.dis;
                e = tmp1.num;
            }
        }
    }

    ma = front = rear = 0;
    clr(vis);
    tmp.num = e;
    tmp.dis = 0;
    que[rear++] = tmp;
    vis[e] = true;
    while(front < rear) {
        tmp = que[front++];
        tmps = tmp.num;
        for(int i = head[tmps]; i; i = edge[i].next)
        if(!vis[edge[i].to]) {
            tmp1.num = edge[i].to;
            tmp1.dis = edge[i].w + tmp.dis;
            vis[edge[i].to] = true;
            que[rear++] = tmp1;
            pre[tmp1.num] = tmps;
            if(ma < tmp1.dis) {
                s = tmp1.num;
                ma = tmp1.dis;
            }
        }
    }

    ma = front = rear = 0;
    clr(vis);
    tmp.num = s;
    tmp.dis = 0;
    que[rear++] = tmp;
//printf("%d===>", s);
    while(pre[s]) {
//printf("%d==>", pre[s]);
        tmp.num = pre[s];
        tmp.dis = 0;
        que[rear++] = tmp;
        vis[s] = true;
        s = pre[s];
    }
//printf("%d", e);
//puts("");
    vis[e] = true;
    while(front < rear) {
        tmp = que[front++];
        tmps = tmp.num;
        for(int i = head[tmps]; i; i = edge[i].next)
        if(!vis[edge[i].to]) {
            tmp1.num = edge[i].to;
            tmp1.dis = edge[i].w + tmp.dis;
            que[rear++] = tmp1;
            vis[tmp1.num] = true;
            if(tmp1.dis > ma) {
                ma = tmp1.dis;
            }
        }
    }
    return ma;
}


int main()
{
    int i, j, k;
    int n, a, b;
    int u, v, w;
    while(scanf("%d", &n) != EOF) {
        if(n == 0) break;

        tot = 1;
        clr(head);
        for(i = 0; i < n-1; ++i) {
            scanf("%d %d %d", &u, &v, &w);
            add(u, v, w);
        }

        printf("%d\n", bfs());
    }
    return 0;
}

题目链接：http://acm.hrbust.edu.cn/index.php?m=ProblemSet&a=showProblem&problem_id=1460