POJ 3268 Silver Cow Party

最短路水题，正向边，反向边各跑一次。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;

const int INF=0x7FFFFFFF;
const int maxn=1000+10;
int n,m,x;
int dis1[maxn],dis2[maxn];
struct Edge
{
    int to,val;
}e1[110000],e2[110000];
int tot;
vector<int>g1[maxn],g2[maxn];
int flag[maxn];

void SPFA1()
{
    for(int i=1;i<=n;i++) dis1[i]=INF;
    queue<int>Q;
    memset(flag,0,sizeof flag);
    dis1[x]=0,flag[x]=1; Q.push(x);
    while(!Q.empty())
    {
        int head=Q.front(); Q.pop(); flag[head]=0;
        for(int i=0;i<g1[head].size();i++)
        {
            int id=g1[head][i];
            if(dis1[head]+e1[id].val<dis1[e1[id].to])
            {
                dis1[e1[id].to]=dis1[head]+e1[id].val;
                if(flag[e1[id].to]==0)
                {
                    flag[e1[id].to]=1;
                    Q.push(e1[id].to);
                }
            }
        }
    }
}

void SPFA2()
{
    for(int i=1;i<=n;i++) dis2[i]=INF;
    queue<int>Q;
    memset(flag,0,sizeof flag);
    dis2[x]=0,flag[x]=1; Q.push(x);
    while(!Q.empty())
    {
        int head=Q.front(); Q.pop(); flag[head]=0;
        for(int i=0;i<g2[head].size();i++)
        {
            int id=g2[head][i];
            if(dis2[head]+e2[id].val<dis2[e2[id].to])
            {
                dis2[e2[id].to]=dis2[head]+e2[id].val;
                if(flag[e2[id].to]==0)
                {
                    flag[e2[id].to]=1;
                    Q.push(e2[id].to);
                }
            }
        }
    }
}

int main()
{
    scanf("%d%d%d",&n,&m,&x); tot=0;
    for(int i=1;i<=m;i++)
    {
        int u,v,w; scanf("%d%d%d",&u,&v,&w);

        tot++;

        e1[tot].to=v,e1[tot].val=w;
        g1[u].push_back(tot);

        e2[tot].to=u,e2[tot].val=w;
        g2[v].push_back(tot);
    }

    SPFA1();
    SPFA2();

    int ans=0;

    for(int i=1;i<=n;i++)
    {
        if(dis1[i]==INF||dis2[i]==INF) continue;
        if(i==x) continue;
        ans=max(ans,dis1[i]+dis2[i]);
    }
    printf("%d\n",ans);

    return 0;
}