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\n",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\n",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\n");
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;
}