poj 1511 Invitation Cards 静态邻接表的SPFA

//用vector写邻接表无情的TLE，只好用静态邻接表，两次SPFA分别求去和回的最短路

#include<iostream>

#include<cstdio>

#include<cstring>

#include<queue>

#include<vector>

using namespace std;

const int INF=0x3f3f3f3f;

const int maxn=1000005;

int n, m, d[maxn], pre[2][maxn], inqueue[maxn], cnt;

struct node{

int v, w, next;

}edge[2][maxn];

void add_edge(int s, int e, int w){

edge[0][cnt].v=e;

edge[0][cnt].w=w;

edge[0][cnt].next=pre[0][s];

pre[0][s]=cnt;

edge[1][cnt].v=s;

edge[1][cnt].w=w;

edge[1][cnt].next=pre[1][e];

pre[1][e]=cnt;

cnt++;

}

void init(){

int i, x, y, c;

cnt=1;

scanf("%d%d", &n, &m);

for(i=1; i<=n; i++){

pre[0][i]=-1;

pre[1][i]=-1;

}

for(i=0; i<m; i++){

scanf("%d%d%d", &x, &y, &c);

add_edge(x, y, c);

}

}

bool spfa(int g, int st){

int i, head, s;

for(i=1; i<=n; i++){

inqueue[i]=0;

d[i]=INF;

}

queue<int > q;

d[st]=0;

inqueue[st]=1;

q.push(st);

while(!q.empty()){

head=q.front();

q.pop();

inqueue[head]=0;

s=pre[g][head];

while(s!=-1){

if(d[edge[g][s].v]>d[head]+edge[g][s].w){

d[edge[g][s].v]=d[head]+edge[g][s].w;

if(!inqueue[edge[g][s].v]){

inqueue[edge[g][s].v]=1;

q.push(edge[g][s].v);

}

}

s=edge[g][s].next;

}

}

return true;

}

int main(){

//freopen("1.txt", "r", stdin);

int i, j, T;

long long ans;

scanf("%d", &T);

while(T--){

init();

ans=0;

for(i=0; i<2; i++){

spfa(i, 1);

for(j=2; j<=n; j++)

ans+=d[j];

}

printf("%lld\n", ans);

}

return 0;

}