CodeForces954DFight Against Traffic（最短路）

Posted 2020-10-28 void_f

Description

题目链接

Solution

从起点和终点分别做一次最短路并记录结果

枚举每一条可能的边判断

Code

#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
#define N 1010
using namespace std;

struct info{int to,nex;}e[N*2];
int n,m,s,t,tot,head[N],dis[N],ddis[N],Ans;
bool g[N][N];

inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

inline void Link(int u,int v){
    e[++tot].to=v;e[tot].nex=head[u];head[u]=tot;
}

bool vis[N];
void bfs(int d[],int s){
    memset(vis,0,sizeof(vis));
    queue<int> q;
    q.push(s);
    d[s]=0,vis[s]=1;
    while(!q.empty()){
        int u=q.front();q.pop();
        for(int i=head[u];i;i=e[i].nex){
            int v=e[i].to;
            if(vis[v]) continue;
            d[v]=d[u]+1;
            q.push(v);
            vis[v]=1;
        }
    }
}

int main(){
    n=read(),m=read(),s=read(),t=read();
    while(m--){int u=read(),v=read();Link(u,v);Link(v,u);g[u][v]=g[v][u]=1;}
    bfs(dis,s);
    bfs(ddis,t);
    for(int i=1;i<=n;++i)
        for(int j=i+1;j<=n;++j)
            if(!g[i][j]&&dis[i]+ddis[j]+1>=dis[t]&&dis[j]+ddis[i]+1>=dis[t])
                Ans++;
    printf("%d\n",Ans);
    return 0;
}