POJ 3259 Wormholes (判负环)

• 题意: 无向图加负的有向边,求是否存在负环
• 思路:
  1. spfa: spfa是一直拿更新后的边来增广其他可达边,如果存在负环,则会一直增广,存下当前路径的长度,如果超过了n则说明一直在负环上增广,即存在负环
  2. floyd: 跑一遍Floyd,如果存在某个点到自己的距离为负,则存在负环(负环上的点走一遍负环,到自己的距离肯定为负)

spfa:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<algorithm>
#define ll long long
#define FOR(i,l,r) for(int i = l ; i <= r ;++i )
#define inf 0x3f3f3f3f
#define EPS (1e-9)
#define ALL(T)  T.begin(),T.end()
using namespace std;

const int maxn = 5010;

int n,m,w;

struct Edge{
	int to,next,w;
}edge[maxn*10];
int tot;
int d[maxn],cnt[maxn],head[maxn];
int v[maxn];

void addEdge(int u,int v,int w){
	edge[tot].to = v;
	edge[tot].w = w;
	edge[tot].next = head[u];
	head[u] = tot++; 
}

bool spfa(){
	queue<int> q;
	memset(v,0,sizeof(v));
	memset(cnt,0,sizeof(cnt));
	d[1] = 0;
	cnt[1] = 0; 
	v[1] = 1;
	q.push(1);
	while(q.size()){
		int x = q.front();	q.pop();
		v[x] = 0;
		for(int i=head[x];i!=-1;i=edge[i].next){
			int y = edge[i].to;
			int z = edge[i].w;
			if(d[y]>d[x]+z){
				d[y] = d[x]+z;
				cnt[y] = cnt[x] + 1;
				if(cnt[y]>n)
					return true;
				if(v[y]==0){
					q.push(y);
					v[y] =1;
					// cnt[y]++;
				}	
			}
		}
	}
	return false;
}


int main(){
int t;
cin >> t;
while(t--){
	cin >> n >> m >> w;
	tot = 0;
	int f,to,wet;
	memset(head,-1,sizeof(head));
	memset(d,inf,sizeof(d));
	for(int i=1;i<=m;++i){
		cin >> f >> to >> wet;
		addEdge(f,to,wet);
		addEdge(to,f,wet);
	}
	for(int i=1;i<=w;++i){
		cin >> f >> to >> wet;
		addEdge(f,to,-wet);
	}
	if(spfa())cout << "YES" << endl;
	else cout <<"NO" << endl;
}
	
	return 0;
}

floyd:

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

const int maxn = 510;

inline int read(){
   int s=0,w=1;
   char ch=getchar();
   while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
   while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
   return s*w;
}

int n,m,w;
int g[maxn][maxn];
inline int min(int a,int b){
	return a>b? b:a;
}
bool floyd(){
	for(int k=1;k<=n;++k){
		for(int i=1;i<=n;++i){
			for(int j=1;j<=n;++j){
				if(g[i][j]>g[i][k]+g[k][j])
					g[i][j] =  g[i][k]+g[k][j];
			}
			if(g[i][i]<0){
				return true;
			}
		}
	}
	return false;
}

int main(){
int t;
t = read();
while(t--){
	n = read();	m = read();	w = read();
	memset(g,0x3f,sizeof(g));
	for(int i=1;i<=n;++i)	g[i][i] = 0;
	int f,to,wet;
	for(int i=1;i<=m;++i){
		f = read();	to = read();	wet = read();
		g[f][to] = wet<g[f][to] ? wet:g[f][to];
		g[to][f] = g[f][to];
	}
	for(int i=1;i<=w;++i){
		f = read();	to = read();	wet = read();
		g[f][to] = -wet;
	}

	if(floyd())printf("YES
");
	else printf("NO
");
}
	
	return 0;
}

floyd加快读 1.6s用min会T,用三目运算符也会T,用if才能过神奇, spfa加cin 0.5s

题目连接