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Posted 2022-12-20 nent

题意：给出一张有向图，求原点到所有点的最短路和所有点到原点的最短路的距离之和

解法：反向建图+spfa

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int MAXP=1000005,MAXQ=1000005;
struct Edgeint next,to,dis;edge[MAXQ][2]; 
int n,p,q,head[MAXP][2],n_e[2],dis[MAXP][2];
bool vis[MAXP][2];

void spfa(int bgn,int x);
void init();
void addedge(int from,int to,int dis,int x);
int main()

    scanf("%d",&n);
    for(int k=1;k<=n;k++)
    
        int ans=0,u,v,w;
        init();
        scanf("%d%d",&p,&q);
        for(int i=1;i<=q;i++)
        
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w,1);
            addedge(v,u,w,0);
        
        spfa(1,1);
        spfa(1,0);
        for(int i=2;i<=p;i++)
            ans+=dis[i][1]+dis[i][0];
        printf("%d\n",ans);
    
    return 0;

void addedge(int from,int to,int dis,int x)

    edge[++n_e[x]][x].next=head[from][x];
    edge[n_e[x]][x].to=to;
    edge[n_e[x]][x].dis=dis;
    head[from][x]=n_e[x];

void init()

    memset(vis,0,sizeof(vis));
    memset(head,0,sizeof(head));
    memset(edge,0,sizeof(edge));
    memset(dis,0x7f,sizeof(dis));


void spfa(int bgn,int x)

    int u,v;
    queue<int> q;
    q.push(bgn); dis[bgn][x]=0; vis[bgn][x]=1;
    while(!q.empty())
    
        u=q.front();
        q.pop();
        vis[u][x]=0;
        for(int i=head[u][x];i;i=edge[i][x].next)
        
            v=edge[i][x].to;
            if(dis[v][x]>dis[u][x]+edge[i][x].dis)
            
                dis[v][x]=dis[u][x]+edge[i][x].dis;
                if(!vis[v][x])
                
                    vis[v][x]!=vis[v][x];
                    q.push(v);