POJ 1985--Cow Marathon【树的直径 && 模板】

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Description

Input

Output

Sample Input

7 6
1 6 13 E
6 3 9 E
3 5 7 S
4 1 3 N
2 4 20 W
4 7 2 S

Sample Output

52

Hint

第一道树的直径题

#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#define maxn 40000+10
#define maxm 80000+10
#define INF 0x3f3f3f3f
using namespace std;

struct node{
    int u, v, w, next;
};

node edge[maxm];
int head[maxn], cnt;
int dist[maxn], vis[maxn];
int n, m;
int sum, ed ;

void init(){
    cnt = 0;
    memset(head, -1, sizeof(head));
}

void add(int u, int v, int w){
    edge[cnt] = {u, v, w, head[u]};
    head[u] = cnt++;
    edge[cnt] = {v, u, w, head[v]};
    head[v] = cnt++;
}

void getmap(){
    char str[5];
    while(m--){
        int a, b, c;
        scanf("%d%d%d%s", &a, &b, &c, str);
        add(a, b, c);
    }
}

void bfs(int sx){
    queue<int>q;
    for(int i = 1; i <= n; ++i){
        vis[i] = 0;
        dist[i] = 0;
    }
    vis[sx] = 1;
    sum = 0;
    ed = sx;
    q.push(sx);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        for(int i = head[u]; i != -1 ;i = edge[i].next){
            int v = edge[i].v;
            if(!vis[v]){
                dist[v] = dist[u] + edge[i].w;
                if(sum < dist[v]){
                    ed = v;
                    sum = dist[v];
                }
                vis[v] = 1;
                q.push(v);
            }
        }
    }
}

int main (){
    while(scanf("%d%d", &n, &m) != EOF){
        init();
        getmap();
        bfs(1);
        bfs(ed);
        printf("%d\n", sum);
    }
    return 0;
}

POJ 1985--Cow Marathon【树的直径 && 模板】

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原文地址：http://blog.csdn.net/hpuhjh/article/details/47728205