poj 1511 Invitation Cards dijkstra+heap

最近没有状态，不太难得一题，TLE了3次，WA了1次。

这题主要就是要正向，逆向两次dijkstra,因为稀疏图，所以用heap优化有明显作用。

注意会超出int范围，要用long long

代码太挫了，越写越挫，估计到瓶颈期了

/*
author:jxy
lang:C/C++
university:China,Xidian University
**If you need to reprint,please indicate the source**
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#define INF 1E9
using namespace std;
long long d[1000001];
bool vis[1000001];
int n;
int u[1000001],va[1000001];
int first[1000001],next[1000001],cnt,m;
int vf[1000001],vnext[1000001],vu[1000001],vvalue[1000001],vcnt;
int T;
struct node
{
int v,x;
friend bool operator <(node a,node b)
{
return a.v>b.v;
}
node(int a,int b)
{
v=a,x=b;
}
node()
{
}
};
void add(int vn,int un,int t)
{
next[cnt]=first[vn];
u[cnt]=un;
va[cnt]=t;
first[vn]=cnt++;

vnext[vcnt]=vf[un];
vu[vcnt]=vn;
vvalue[vcnt]=t;
vf[un]=vcnt++;
}
priority_queue<node> q;
void dijkstra(int x)
{
memset(d,127,(n+1)*sizeof(d[0]));
memset(vis,0,(n+1)*sizeof(vis[0]));
while(!q.empty())q.pop();
d[x]=0;
q.push(node(d[x],x));
int i,t,v;
while(!q.empty())
{
v=q.top().x;q.pop();
if(vis[v])continue;
vis[v]=1;
for(i=first[v];i!=-1;i=next[i])
{
if(d[u[i]]<=(t=d[v]+va[i]))continue;
d[u[i]]=t;
q.push(node(d[u[i]],u[i]));
}
}
}
void di2(int x)
{
memset(d,127,(n+1)*sizeof(d[0]));
memset(vis,0,(n+1)*sizeof(vis[0]));
while(!q.empty())q.pop();
d[x]=0;
q.push(node(d[x],x));
int i,t,v;
while(!q.empty())
{
v=q.top().x;q.pop();
if(vis[v])continue;
vis[v]=1;
for(i=vf[v];i!=-1;i=vnext[i])
{
if(d[vu[i]]<=(t=d[v]+vvalue[i]))continue;
d[vu[i]]=t;
q.push(node(d[vu[i]],vu[i]));
}
}
}
int main()
{
int i,a,b,t;
scanf("%d",&T);
memset(next,-1,sizeof(next));
memset(first,-1,sizeof(first));
memset(vnext,-1,sizeof(vnext));
memset(vf,-1,sizeof(vf));
while(T--)
{
scanf("%d%d",&n,&m);
cnt=0;
for(i=0;i<m;i++)
{
scanf("%d%d%d",&a,&b,&t);
add(a,b,t);
}
long long ans=0;
dijkstra(1);
for(i=2;i<=n;i++) ans+=d[i];
di2(1);
for(i=2;i<=n;i++) ans+=d[i];
printf("%lld\n",ans);
memset(next,-1,(m+1)*sizeof(next[0]));
memset(first,-1,(n+1)*sizeof(first[0]));
memset(vnext,-1,(m+1)*sizeof(vnext[0]));
memset(vf,-1,(n+1)*sizeof(vf[0]));
}
}