简介
对于一个不存在负环的图，从起点到任意一个点最短距离经过的点最多只有 n 个。用 cnt[i] 表示从起点（假设是 1）到i的最短距离包含点的个数，初始化 cnt[1]=1，那么当我们能够用点u松弛点v时，松弛时同时更新cnt[v] = cnt[u]+1，若发现此时 cnt[v] > n，那么就存在负环
还有一种方法是记录每个点的入队次数，入队次数大于 n 就说明有负环，但是这样做一般都要比上面的方法慢。举个例子，在一个由 n 个点构成的负环中，这个方法要绕环n次，而上面的方法绕环 1 次就行了
例题
Luogu-P3385
模板题
第一种方案
#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<cstring>
#include<algorithm>
#define lson x<<1
#define rson x<<1|1
#define ll long long
#define rint register int
#define mid ((L + R) >> 1)
using namespace std;
template <typename xxx> inline void read(xxx &x) {
char c = getchar(),f = 1;x = 0;
for(;c ^ '-' && !isdigit(c);c = getchar());
if(c == '-') c = getchar(),f = -1;
for(;isdigit(c);c = getchar()) x = (x<<1) + (x<<3) + (c ^ '0');
x *= f;
}
template<typename xxx>void print(xxx x)
{
if(x<0){putchar('-');x=-x;}
if(x>9) print(x/10);
putchar(x%10+'0');
}
const int maxn = 100010;
const int inf = 0x7fffffff;
const int mod = 1e9 + 7;
struct edge {
int to,last,val;
}e[maxn];
int head[maxn],tot;
inline void add(int from,int to,int val) {
++tot;
e[tot].to = to;
e[tot].val = val;
e[tot].last = head[from];
head[from] = tot;
}
int n,m;
int dis[maxn],vis[maxn],tim[maxn];
inline int spfa() {
queue<int>q;
for(rint i = 0;i <= n; ++i) {
dis[i] = inf;
vis[i] = 0;
tim[i] = 0;
}
q.push(1);dis[1] = 0;tim[1] = 1;
while(q.size()) {
int x = q.front();q.pop();vis[x] = 0;
for(rint i = head[x];i;i = e[i].last) {
if(dis[e[i].to] > dis[x] + e[i].val) {
dis[e[i].to] = dis[x] + e[i].val;
tim[e[i].to] = tim[x] + 1;
if(tim[e[i].to] > n) return 1;
if(!vis[e[i].to]) {
vis[e[i].to] = 1;
q.push(e[i].to);
}
}
}
}
return 0;
}
int main() {
int t;read(t);
while(t--) {
tot = 0;
memset(head,0,sizeof(head));
read(n);read(m);
for(rint i = 1;i <= m; ++i) {
int a,b,c;
read(a);read(b);read(c);
add(a,b,c);
if(c >= 0) add(b,a,c);
}
if(spfa()) printf("YE5\n");
else printf("N0\n");
}
return 0;
}
/*
*/第二种方案(实测着实慢很多)
#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<cstring>
#include<algorithm>
#define lson x<<1
#define rson x<<1|1
#define ll long long
#define rint register int
#define mid ((L + R) >> 1)
using namespace std;
template <typename xxx> inline void read(xxx &x) {
char c = getchar(),f = 1;x = 0;
for(;c ^ '-' && !isdigit(c);c = getchar());
if(c == '-') c = getchar(),f = -1;
for(;isdigit(c);c = getchar()) x = (x<<1) + (x<<3) + (c ^ '0');
x *= f;
}
template<typename xxx>void print(xxx x)
{
if(x<0){putchar('-');x=-x;}
if(x>9) print(x/10);
putchar(x%10+'0');
}
const int maxn = 100010;
const int inf = 0x7fffffff;
const int mod = 1e9 + 7;
struct edge {
int to,last,val;
}e[maxn];
int head[maxn],tot;
inline void add(int from,int to,int val) {
++tot;
e[tot].to = to;
e[tot].val = val;
e[tot].last = head[from];
head[from] = tot;
}
int n,m;
int dis[maxn],vis[maxn],tim[maxn];
inline int spfa() {
queue<int>q;
for(rint i = 0;i <= n; ++i) {
dis[i] = inf;
vis[i] = 0;
tim[i] = 0;
}
q.push(1);dis[1] = 0;
while(q.size()) {
int x = q.front();q.pop();vis[x] = 0;
if(tim[x] == n) return 1;
++tim[x];
for(rint i = head[x];i;i = e[i].last) {
if(dis[e[i].to] > dis[x] + e[i].val) {
dis[e[i].to] = dis[x] + e[i].val;
if(!vis[e[i].to]) {
vis[e[i].to] = 1;
q.push(e[i].to);
}
}
}
}
return 0;
}
int main() {
// freopen("my.out","w",stdout);
int t;read(t);
while(t--) {
tot = 0;
memset(head,0,sizeof(head));
read(n);read(m);
for(rint i = 1;i <= m; ++i) {
int a,b,c;
read(a);read(b);read(c);
add(a,b,c);
if(c >= 0) add(b,a,c);
}
if(spfa()) printf("YE5\n");
else printf("N0\n");
}
return 0;
}
/*
*/