题解:最短路计数

洛谷P1144

Posted by TH911 on January 23, 2025

题目传送门

题意分析

因为是无向无权图,因此可以通过 BFS 求最短路。

BFS 找到即最优,因此找的时候递推即可。

AC 代码

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//#include<bits/stdc++.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<iomanip>
#include<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<ctime>
#include<deque>
#include<queue>
#include<stack>
#include<list>
using namespace std;
const int N=1e6,M=2e6,mod=100003;
struct edge{
	int v,r;
}a[2*M+1];
int n,m,u,v,h[N+1],ans[N+1];
void create(int u,int v){
	static int top=0;
	a[++top]={v,h[u]};
	h[u]=top;
}
void bfs(int s){
	static int pl,dis[N+1];
	queue<int>q;
	q.push(s);
	dis[s]=1;
	ans[s]=1;
	while(q.size()){
		pl=q.front();
		q.pop();
		for(int i=h[pl];i>0;i=a[i].r){
			if(dis[a[i].v]==0){
				q.push(a[i].v);
				dis[a[i].v]=dis[pl]+1;
				ans[a[i].v]=ans[pl];
			}else if(dis[pl]+1==dis[a[i].v])ans[a[i].v]+=ans[pl];
			ans[a[i].v]%=mod;
		}
	}
}
int main(){
	/*freopen("test.in","r",stdin);
	freopen("test.out","w",stdout);*/
	
	scanf("%d %d",&n,&m);
	for(int i=1;i<=m;i++){
		scanf("%d %d",&u,&v);
		create(u,v);create(v,u);
	}bfs(1);
	for(int i=1;i<=n;i++)printf("%d\n",ans[i]);
	
	/*fclose(stdin);
	fclose(stdout);*/
	return 0;
}