# HDU 1535 Invitation Cards （POJ 1511）

d1[]为 1~N 的最短路。

d2[] 为 1~N 的最短路。

POJ 数据也真是屌。完全不看题意的。

```#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<queue>
#include<map>
#include<stack>
#include<iostream>
#include<list>
#include<set>
#include<cmath>
#define INF 0x7fffffff
#define eps 1e-6
using namespace std;
int n,m;
struct lx
{
int v,d;
};
int dis[1000001];
bool vis[1000001];
int e[1000001];
vector<lx>g[1000001];
void swapg()
{
for(int i=1;i<=n;i++)
e[i]=g[i].size();
for(int i=1;i<=n;i++)
{
int u,v,d;
lx now;
u=i;
for(int j=0;j<e[i];j++)
{
v=g[u][j].v,d=g[u][j].d;
now.d=d,now.v=u;
g[v].push_back(now);
}
}
}
int SPFA(int thend) // long long
{
for(int i=1;i<=n;i++)
dis[i]=INF,vis[i]=0;
queue<int>q;
dis[1]=0,vis[1]=1;
q.push(1);
while(!q.empty())
{
int u=q.front();q.pop();
vis[u]=0;
for(int j=e[u];j<g[u].size();j++)
{
int v=g[u][j].v;
int d=g[u][j].d;
if(dis[v]>dis[u]+d)
{
dis[v]=dis[u]+d;
if(!vis[v])
{
vis[v]=1;
q.push(v);
}
}
}
}
int ans=0; //long long
for(int i=1;i<=n;i++)
ans+=dis[i];
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
int u,v,d;
lx now;
for(int i=1;i<=n;i++)
g[i].clear();
while(m--)
{
scanf("%d%d%d",&u,&v,&d);
now.d=d;
now.v=v;
g[u].push_back(now);
}
memset(e,0,sizeof(e));
int dis1=SPFA(n); //long long
swapg();
int dis2=SPFA(n); //long long
printf("%d\n",dis1+dis2); // lld%
}
}
```

HDU 1535 Invitation Cards （POJ 1511）,布布扣,bubuko.com

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