POJ 1511 Invitation Cards （spfa）

来源：互联网发布：java就业班编辑：程序博客网时间：2024/05/03 00:51

求到某点的来回最短距离，数据太大，要用spfa邻接矩阵存储，将两点互换再求一次最短路，所有距离和即为所求

#include "string"#include "iostream"#include "cstdio"#include "cmath"#include "set"#include "queue"#include "vector"#include "cctype"#include "sstream"#include "cstdlib"#include "cstring"#include "stack"#include "ctime"#include "algorithm"#define pa pair<int,int>#define Pi M_PI#define INF 0x3f3f3f3f#define INFL 0x3f3f3f3f3f3f3f3fLLusing namespace std;typedef long long LL;const int M=1000005;struct node{int e,next,w;}edge[2][M];LL d[M],ans;int n,m,headlist[2][M],vis[M];void spfa(int t){int v,i,b;queue <int>q;    for(i=1;i<=n;i++){d[i]=INF;vis[i]=0;}q.push(1);vis[1]=1;d[1]=0;while(!q.empty()){v=q.front();q.pop();vis[v]=0;for(i=headlist[t][v];i!=-1;i=edge[t][i].next){b=edge[t][i].e;if(d[b]>d[v]+edge[t][i].w){d[b]=d[v]+edge[t][i].w;if(!vis[b]){vis[b]=1;q.push(b);}}}}}int main(){int t,i,a,b,w;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);for(i=1;i<=n;i++){headlist[0][i]=-1;headlist[1][i]=-1;}for(i=0;i<m;i++) {scanf("%d%d%d",&a,&b,&w);      //数组模拟邻接表 edge[0][i].w=w;edge[0][i].e=b;edge[0][i].next=headlist[0][a]; headlist[0][a]=i;edge[1][i].w=w;edge[1][i].e=a;edge[1][i].next=headlist[1][b];headlist[1][b]=i;}ans=0;//构造完图，直接模板 spfa(0);for(i=1;i<=n;i++) ans+=d[i];spfa(1);for(i=1;i<=n;i++) ans+=d[i];printf("%I64d\n",ans);}return 0;}

0 0