URAL 1227 Rally Championship（树的直径）（无向图判环）

1227. Rally Championship

Time limit: 1.0 second
Memory limit: 64 MB

A high-level international rally championship is about to be held. The rules of the race state that the race is held on ordinary roads and the route has a fixed length. You are given a map of the cities and two-way roads connecting it. To make the race safer it is held on one-way roads. The race may start and finish anyplace on the road. Determine if it is possible to make a route having a given length S.

Input

The first line of the input contains integers M, N and S that are the number of cities, the number of roads the length of the route (1 ≤ M ≤ 100; 1 ≤ N ≤ 10 000; 1 ≤ S ≤ 2 · 10⁶).

The following N lines describe the roads as triples of integers: P, Q, R. Here P and Q are cities connected with a road, and R is the length of this road. All numbers satisfy the following restrictions: 1 ≤ P, Q ≤ M; 1 ≤ R ≤ 32000.

Output

Write YES to the output if it is possible to make a required route and NO otherwise. Note that answer must be written in capital Latin letters.

Samples

input	output
3 2 20 1 2 10 2 3 5	NO
3 3 1000 1 2 1 2 3 1 1 3 1	YES

Problem Source: 2002-2003 ACM Central Region of Russia Quarterfinal Programming Contest, Rybinsk,

【分析】先判断是否有环，如果有则YES，没有的话找最大直径，如果大于等于s，则YES，其他情况则NO。注意有图不连通情况。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
typedef long long ll;
using namespace std;
const int N = 10005;
const int M = 24005;
int n,m,cnt=0;
int tot=0,s,t,son,sum;
int head[N],dis[N],vis[N],pre[N],vis1[N];
int w[N][N];
int in[N],out[N];
int  bfs(int x) {
    met(vis,0);
    met(dis,0);
    sum=0;
    queue<int>Q;
    Q.push(x);
    vis[x]=1;vis1[x]=1;
    while(!Q.empty()) {
        int t=Q.front();
        Q.pop();//printf("t=%d
",t);
        bool has=false;
        for(int i=1; i<=n; i++) {
            if(!vis[i]&&w[t][i]!=0) {
                Q.push(i);
                dis[i]=dis[t]+w[t][i];
                vis[i]=vis1[i]=1;
                has=true;
            }
        }
        if(!has) {
            if(dis[t]>sum) {
                sum=dis[t];
                //printf("%d %d
",sum,dis[t]);
                son=t;
            }
        }
    }
    return sum;
}
int main() {
    int u,v,l,sum=0;
    scanf("%d%d%d",&n,&m,&s);
    while(m--) {
        scanf("%d%d%d",&u,&v,&l);
        w[u][v]=w[v][u]=l;
        in[u]++;
        in[v]++;
    }
    queue<int>q;
    for(int i=1; i<=n; i++) {
        if(in[i]<=1)q.push(i);
    }
    while(!q.empty()) {
        int t=q.front();
        q.pop();
        vis[t]=1;
        for(int i=1; i<=n; i++) {
            if(!vis[i]&&w[t][i]!=0) {
                in[i]--;
                if(in[i]==1)q.push(i);
            }
        }
    }
    bool flag=false;
    for(int i=1; i<=n; i++) {
        if(!vis[i])flag=true;
    }
    if(flag)printf("YES
");
    else {
        int ans=-1;
        memset(vis1,0,sizeof vis1);
        for(int i=1; i<=n; i++)
            if(!vis1[i]) {
                bfs(i);
                ans=max(ans,bfs(son));
            }
        if(ans>=s)puts("YES");
        else puts("NO");
    }
    return 0;
}