最短路 CF954D Fight Against Traffic

CF954D Fight Against Traffic

题意描述：

给你一张无向图,一共有n个点(2 <= n <= 1000),由m条边连接起来(1 <= m <= 10000),现在要在任意一对没有连边的点之间连上一条边,并且保证s到t之间的最短路径长度不变(最短路径长度表示s到t最少经过的边的数量)(1 <= s,t <= n , s≠t),请你求出一共有多少条这样的边。

被水题成功报复。。。

code：

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>

using namespace std;

inline int read(){
	int sum=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();}
	while(ch>='0'&&ch<='9'){sum=(sum<<1)+(sum<<3)+ch-'0'; ch=getchar();}
	return sum*f;
}

const int wx=2017;

int head[wx],dis1[wx],dis2[wx];
int vis[wx],fl[wx][wx];
int num,n,m,s,t,ans;

struct e{
	int nxt,to,dis;
}edge[wx*2];

void add(int from,int to,int dis){
	edge[++num].nxt=head[from];
	edge[num].to=to;
	edge[num].dis=dis;
	head[from]=num;
}

struct node{
	int u,d;
	friend bool operator < (const node & a,const node & b){
		return a.d>b.d;
	}
};

priority_queue<node > q;

void Dij(int S,int dis[]){
	for(int i=1;i<=n;i++)dis[i]=0x3f3f3f3f,vis[i]=0;
	dis[S]=0; q.push((node){S,0});
	while(q.size()){
		int u=q.top().u; q.pop();
		if(vis[u])continue; vis[u]=1;
		for(int i=head[u];i;i=edge[i].nxt){
			int v=edge[i].to;
			if(dis[v]>dis[u]+edge[i].dis){
				dis[v]=dis[u]+edge[i].dis;
				q.push((node){v,dis[v]});
			}
		}
	}
}

int main(){
	n=read(); m=read();
	s=read(); t=read();
	for(int i=1;i<=m;i++){
		int x,y;
		x=read(); y=read();
		add(x,y,1); add(y,x,1); 
		fl[x][y]=1; fl[y][x]=1;
	}
	Dij(s,dis1); Dij(t,dis2);
	for(int i=1;i<=n;i++){
		for(int j=i+1;j<=n;j++){
			if(fl[i][j])continue;
			int tmp1=dis1[i]+dis2[j]+1;
			int tmp2=dis1[j]+dis2[i]+1;
			if(min(tmp1,tmp2)>=dis1[t])ans++;
		}
	}
	printf("%d
",ans);
	return 0;
}