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

2

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

2

思路 :

剪完图后,记录一个最小边权的个数,如果从1到v的距离有多个,那么就可以考虑删除v

代码

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#define int long long
#define maxn 1100000
using namespace std ;
struct dy{int x , y , z , next ;
}a[maxn] ;
int head[maxn] , vis[maxn] , dis[maxn] ;
int n , m ,k  ;
int U[maxn] , V[maxn] ,in[maxn] ;
typedef pair<int,int>pa ;
priority_queue<pa,vector<pa>,greater<pa> >q ;
int t ;
void add(int x , int y , int z) {a[++t].x = x ;a[t].y = y ;a[t].z = z ;a[t].next = head[x] ;head[x] = t ;
}
void dij(int s) {memset(dis,0x3f,sizeof(dis)) ;memset(vis,0,sizeof(vis)) ;dis[s] = 0 , vis[s] = 0 ;q.push(make_pair(dis[s],s)) ;while(!q.empty()) {int u = q.top().second ;q.pop() ;if(vis[u]) continue ;vis[u] = 1 ;for(int i = head[u] ; i ; i = a[i].next ) {int v = a[i].y ;if(dis[v] > dis[u] + a[i].z ) {dis[v] = dis[u] + a[i].z ;in[v] = 1 ;q.push(make_pair(dis[v],v)) ;}else if(dis[v] == dis[u] + a[i].z) {in[v] ++ ;}}}
}
signed main () {scanf("%lld%lld%lld",&n,&m,&k) ;while(m --) {int x , y , z ;scanf("%lld%lld%lld",&x,&y,&z) ;add(x,y,z) ;add(y,x,z) ;}for(int i = 1 ; i <= k ; i ++) {scanf("%lld%lld",&U[i],&V[i]) ;add(1,U[i],V[i]) ;add(U[i],1,V[i]) ;}dij(1) ;int ans = 0 ;for(int i = 1 ; i <= k ; i ++) {if(dis[U[i]] < V[i] ) ans ++ ;else {if(dis[U[i]] == V[i] && in[U[i]] > 1) ans ++ , in[U[i]] -- ;}}printf("%lld\n",ans) ;return 0 ;
}