spfa

7.bellman-ford


有边数限制的最短路

![在这里插入图片描述]( https://img-blog.csdnimg.cn/20210518203136576.png?x-oss-process=image/watermark ,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl81MzAzNzM3OQ==,size_16,color_FFFFFF,t_70)

输入样例:

3 3 1
1 2 1
2 3 1
1 3 3

输出样例:

3

模板:

//有边数限制的最短路 
//Bellman-Fort 
//单源最短路径存在负权边
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N = 510,M = 100010;

int n,m,k;
int dist[N],backup[N];
struct Edge{
	int a,b,w;
}edges[N];
int bellman_Fort(){
	memset(dist,0x3f,sizeof dist);
	dist[1]=0;
	
	for(int i=0;i<k;i++){
		memcpy(backup,dist,sizeof dist);
		for(int j=0;j<m;j++){
			int a=edges[j].a,b=edges[j].b,w=edges[j].w;  //松弛操作 
			dist[b]=min(dist[b],backup[a]+w);            //三角不等式 
		}
	}
	if(dist[n]>0x3f3f3f3f/2) return -1;
	return dist[n];
}
int main(){
	scanf("%d%d%d",&n,&m,&k);
	
	for(int i=0;i<m;i++){
		int a,b,w;
		scanf("%d%d%d",&a,&b,&w);
		edges[i]={a,b,w};
	}
	
	int t=bellman_Fort();
	
	if(t==-1) puts("impossible");
	else printf("%d\n",t);
	
	return 0;
}

 

8.spfa


spfa求最短路

![在这里插入图片描述]( https://img-blog.csdnimg.cn/2021051820362189.png?x-oss-process=image/watermark ,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl81MzAzNzM3OQ==,size_16,color_FFFFFF,t_70)
时间复杂度 (O)m
输入样例:

3 3
1 2 5
2 3 -3
1 3 4

输出样例:

2

模板:

//最短路 SPFA
//单源最短路径存在负权边
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int N = 100010;
typedef pair<int,int> PII;
int n,m;
int h[N],e[N],ne[N],w[N],idx;
int dist[N];
bool st[N];
void add(int a,int b,int c){
	e[idx]=b;
	w[idx]=c;
	ne[idx]=h[a];
	h[a]=idx++;
}

int spfa(){
	memset(dist,0x3f,sizeof dist);
	dist[1]=0;
	
	queue<int> q;
	q.push(1);
	st[1]=true;
	
	while(q.size()){
		int t=q.front();
		q.pop();
		
		st[t]=false;
		for(int i=h[t];i!=-1;i=ne[i]){
			int j=e[i];
			if(dist[j]>dist[t]+w[i]){
				dist[j]=dist[t]+w[i];
				if(!st[j]){ //入队时判重
					q.push(j);
					st[j]=true;
				}
			}
		}
	}
	
	if(dist[n]==0x3f3f3f3f) return -1;
	return dist[n];
}
int main(){
	scanf("%d%d",&n,&m);
	
	memset(h,-1,sizeof h);
	
	while(m--){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		add(a,b,c);
	}
	
	int t=spfa();
	
	if(t==-1) puts("impossible");
	else printf("%d\n",t);
	
	return 0;
}

 

spfa判断负环

![在这里插入图片描述]( https://img-blog.csdnimg.cn/20210518203719937.png?x-oss-process=image/watermark ,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl81MzAzNzM3OQ==,size_16,color_FFFFFF,t_70)

输入样例:

3 3
1 2 -1
2 3 4
3 1 -4

输出样例:

Yes

模板:

//spfa判断负环 
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int N = 100010;
typedef pair<int,int> PII;
int n,m;
int h[N],e[N],ne[N],w[N],idx;
int dist[N],cnt[N];
bool st[N];
void add(int a,int b,int c){
	e[idx]=b;
	w[idx]=c;
	ne[idx]=h[a];
	h[a]=idx++;
}

bool spfa(){
	queue<int> q;
	for(int i=1;i<=n;i++){
		st[i]=true;
		q.push(i); 
	}
	
	while(q.size()){
		int t=q.front();
		q.pop();
		
		st[t]=false;
		
		for(int i=h[t];i!=-1;i=ne[i]){
			int j=e[i];
			if(dist[j]>dist[t]+w[i]){
				dist[j]=dist[t]+w[i];
				
				cnt[j]=cnt[t]+1;
				if(cnt[j]>=n) return true;
				
				if(!st[j]){
					q.push(j);
					st[j]=true;
				}
			}
		}
	}
	
	return false;
}
int main(){
	scanf("%d%d",&n,&m);
	
	memset(h,-1,sizeof h);
	
	while(m--){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		add(a,b,c);
	}
	
	if(spfa()) puts("Yes");
	else puts("No");
	
	return 0;
}


 

posted @ 2022-03-22 01:06  panse·  阅读(25)  评论(0)    收藏  举报