poj 3268 Silver Cow Party(dijkstra||SPFA)(中等)

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires T_i (1 ≤ T_i ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow‘s return route might be different from her original route to the party since roads are one-way.

Input

Output

Sample Input

4 8 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3

Sample Output

10

题意：

奶牛去目的地的最短路加上返回的最短路，求所有奶牛必须花费的最大时间。

思路：

最短路的算法都可以，要注意设立两个dist数组，分别记录奶牛去的最短路和返回的最短路。最后求dist1[]+dist2[]最大值即。

代码：

//SPFA，其实代码不多，基本一样

//440K	110MS
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
#define maxx 1010
#define inf 0x7fffffff
struct Edge
{
    int to,v;
};
vector<Edge> E1[maxx];
vector<Edge> E2[maxx];
queue<int>q;
bool vis[maxx];//是否入队
int dist[maxx];
int dist1[maxx];
int n,m,x;

void spfa()
{
    memset(vis,false,sizeof(vis));
    while(!q.empty())
        q.pop();
    for(int i=1;i<=n;i++)
        dist[i]=inf;
    dist[x]=0;
    q.push(x);
    vis[x]=true;
    while(!q.empty())
    {
        int temp=q.front();
        for(int i=0;i<(int)E1[temp].size();i++)
        {
            Edge *t=&E1[temp][i];
            if(dist[temp]+(*t).v<dist[(*t).to])
            {
                dist[(*t).to]=dist[temp]+(*t).v;
                if(!vis[(*t).to])
                {
                    q.push((*t).to);
                    vis[(*t).to]=true;
                }
            }
        }
        q.pop();
        vis[temp]=false;
    }

    memset(vis,false,sizeof(vis));
    while(!q.empty())
        q.pop();
    for(int i=1;i<=n;i++)
        dist1[i]=inf;
    dist1[x]=0;
    q.push(x);
    vis[x]=true;
    while(!q.empty())
    {
        int temp=q.front();
        for(int i=0;i<(int)E2[temp].size();i++)
        {
            Edge *t=&E2[temp][i];
            if(dist1[temp]+(*t).v<dist1[(*t).to])
            {
                dist1[(*t).to]=dist1[temp]+(*t).v;
                if(!vis[(*t).to])
                {
                    q.push((*t).to);
                    vis[(*t).to]=true;
                //cnt[(*t).to]++;
                }
            }
        }
        q.pop();
        vis[temp]=false;
    }
    int minn=-1;
    for(int i=1;i<=n;i++)
    {
        if(dist[i]+dist1[i]>minn)
            minn=dist[i]+dist1[i];
    }
    cout<<minn<<endl;
}

int main()
{
    cin>>n>>m>>x;
    for(int i=0,a,b,t;i<m;i++)
    {
        cin>>a>>b>>t;
        Edge tmp;
        tmp.to=b;
        tmp.v=t;
        E1[a].push_back(tmp);
        tmp.to=a;
        tmp.v=t;
        E2[b].push_back(tmp);
    }
    spfa();
    return 0;
}

记的修改一下

#include<iostream>
#include<cstring>
using namespace std;
#define max 1010
#define inf 0x7fffffff
int G[max][max],d[max],db[max],vis[max];
int n,m,x;

void init(int n)
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            G[i][j]=((i==j)?0:inf);
}

int dij()
{
    memset(vis,0,sizeof(vis));
    int min,k;
    for(int i=1;i<=n;i++)
    {
        d[i]=G[x][i];
        db[i]=G[i][x];
    }
    d[x]=0;vis[x]=1;
    db[x]=0;
    for(int i=1;i<n;i++)
    {
        min=inf;
        for(int j=1;j<=n;j++)
            if(!vis[j]&&d[j]<min)
            {
                min=d[j];
                k=j;
            }
        vis[k]=1;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&d[j]>d[k]+G[k][j])
                d[j]=d[k]+G[k][j];
        }
    }
    memset(vis,0,sizeof(vis));
    vis[x]=1;
    for(int i=1;i<n;i++)
    {
        min=inf;
        for(int j=1;j<=n;j++)
            if(!vis[j]&&db[j]<min)
            {
                min=db[j];k=j;
            }
        vis[k]=1;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&db[j]>db[k]+G[k][j])
                db[j]=db[k]+G[k][j];
        }
    }
    int mi=-1;
    for(int i=1;i<=n;i++)
        if(d[i]+db[i]>mi)
          mi=d[i]+db[i];
    return mi;
}

int main()
{
    cin>>n>>m>>x;
    int a,b,c;
    init(n);
    for(int i=1;i<=m;i++)
    {
        cin>>a>>b>>c;
        G[a][b]=c;
    }
    cout<<dij()<<endl;
    return 0;
}