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\n");
    else printf("NO\n");
}
    
    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