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  • POJ-3268

    Silver Cow Party
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 13738   Accepted: 6195

    Description

    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 Ti (1 ≤ Ti ≤ 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.

    Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

    Input

    Line 1: Three space-separated integers, respectively: N, M, and X
    Lines 2..M+1: Line i+1 describes road i with three space-separated integers: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.

    Output

    Line 1: One integer: the maximum of time any one cow must walk.

    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

    Hint

    Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units 
    /**
          题意：最短路中哪个走的路程最大
          解法：SPFA
**/
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#define maxn 1000 + 10
#define INF 0x3f3f3f3f
using namespace std;
int vis[maxn];
int dist[maxn];
int dist1[maxn];
int flag[maxn];
struct Node
{
    int v;
    int cost;
    Node(int _v,int _cost):v(_v),cost(_cost) {}
};
vector<Node>edge[maxn];
vector<Node>edge1[maxn];
int n,m,p;
void addedge(int u,int v,int w)
{
    edge[u].push_back(Node(v,w));
    edge1[v].push_back(Node(u,w));
}
void SPFA(int start)
{
    memset(vis,0,sizeof(vis));
    memset(dist,INF,sizeof(dist));
    queue<int>que;
    que.push(start);
    vis[start] = 1;
    dist[start] = 0;
    while(!que.empty())
    {
        int tt = que.front();
        que.pop();
        vis[tt] = 0;
        for(int i=0; i<edge[tt].size(); i++)
        {
            int mm = edge[tt][i].v;
            if(dist[mm] > dist[tt] + edge[tt][i].cost)
            {
                dist[mm] = dist[tt] + edge[tt][i].cost;
                if(!vis[mm])
                {
                    vis[mm] = 1;
                    que.push(mm);
                }
            }
        }
    }
    memset(vis,0,sizeof(vis));
    memset(dist1,INF,sizeof(dist1));
    while(!que.empty()) que.pop();
    vis[start] = 1;
    dist1[start] = 0;
    que.push(start);
    while(!que.empty())
    {
        int tt = que.front();
        que.pop();
        vis[tt] = 0;
        for(int i=0; i<edge1[tt].size(); i++)
        {
            int mm = edge1[tt][i].v;
            if(dist1[mm] > dist1[tt] + edge1[tt][i].cost)
            {
                dist1[mm] = dist1[tt] + edge1[tt][i].cost;
                if(!vis[mm])
                {
                    vis[mm] = 1;
                    que.push(mm);
                }
            }
        }
    }
    return ;
}
int main()
{
//#ifndef ONLINE_JUDGE
//          freopen("in.txt","r",stdin);
//#endif // ONLINE_JUDGE
    scanf("%d %d %d",&n,&m,&p);
    int u,v,w;
    for(int i=0; i<m; i++)
    {
        scanf("%d %d %d",&u,&v,&w);
        addedge(u,v,w);
    }
    SPFA(p);
    int mmax = -INF;
    for(int i=1; i<=n; i++)
    {
        if(dist[i] + dist1[i] > mmax && dist[i] != INF && dist1[i]!= INF )
            mmax = dist[i] + dist1[i];
    }
    printf("%d
",mmax);
    return 0;
}
    /**
          题意：最短路中哪个走的路程最大
          解法：SPFA
**/
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#define maxn 1000 + 10
#define INF 0x3f3f3f3f
using namespace std;
int vis[maxn];
int dist[maxn];
int dist1[maxn];
int flag[maxn];
struct Node
{
    int v;
    int cost;
    Node(int _v,int _cost):v(_v),cost(_cost) {}
};
vector<Node>edge[maxn];
vector<Node>edge1[maxn];
int n,m,p;
void addedge(int u,int v,int w)
{
    edge[u].push_back(Node(v,w));
    edge1[v].push_back(Node(u,w));
}
void SPFA(int start)
{
    memset(vis,0,sizeof(vis));
    memset(dist,INF,sizeof(dist));
    queue<int>que;
    que.push(start);
    vis[start] = 1;
    dist[start] = 0;
    while(!que.empty())
    {
        int tt = que.front();
        que.pop();
        vis[tt] = 0;
        for(int i=0; i<edge[tt].size(); i++)
        {
            int mm = edge[tt][i].v;
            if(dist[mm] > dist[tt] + edge[tt][i].cost)
            {
                dist[mm] = dist[tt] + edge[tt][i].cost;
                if(!vis[mm])
                {
                    vis[mm] = 1;
                    que.push(mm);
                }
            }
        }
    }
    memset(vis,0,sizeof(vis));
    memset(dist1,INF,sizeof(dist1));
    while(!que.empty()) que.pop();
    vis[start] = 1;
    dist1[start] = 0;
    que.push(start);
    while(!que.empty())
    {
        int tt = que.front();
        que.pop();
        vis[tt] = 0;
        for(int i=0; i<edge1[tt].size(); i++)
        {
            int mm = edge1[tt][i].v;
            if(dist1[mm] > dist1[tt] + edge1[tt][i].cost)
            {
                dist1[mm] = dist1[tt] + edge1[tt][i].cost;
                if(!vis[mm])
                {
                    vis[mm] = 1;
                    que.push(mm);
                }
            }
        }
    }
    return ;
}
int main()
{
//#ifndef ONLINE_JUDGE
//          freopen("in.txt","r",stdin);
//#endif // ONLINE_JUDGE
    scanf("%d %d %d",&n,&m,&p);
    int u,v,w;
    for(int i=0; i<m; i++)
    {
        scanf("%d %d %d",&u,&v,&w);
        addedge(u,v,w);
    }
    SPFA(p);
    int mmax = -INF;
    for(int i=1; i<=n; i++)
    {
        if(dist[i] + dist1[i] > mmax && dist[i] != INF && dist1[i]!= INF )
            mmax = dist[i] + dist1[i];
    }
    printf("%d
",mmax);
    return 0;
}
    /**
          题意：最短路中哪个走的路程最大
          解法：SPFA
**/
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#define maxn 1000 + 10
#define INF 0x3f3f3f3f
using namespace std;
int vis[maxn];
int dist[maxn];
int dist1[maxn];
int flag[maxn];
struct Node
{
    int v;
    int cost;
    Node(int _v,int _cost):v(_v),cost(_cost) {}
};
vector<Node>edge[maxn];
vector<Node>edge1[maxn];
int n,m,p;
void addedge(int u,int v,int w)
{
    edge[u].push_back(Node(v,w));
    edge1[v].push_back(Node(u,w));
}
void SPFA(int start)
{
    memset(vis,0,sizeof(vis));
    memset(dist,INF,sizeof(dist));
    queue<int>que;
    que.push(start);
    vis[start] = 1;
    dist[start] = 0;
    while(!que.empty())
    {
        int tt = que.front();
        que.pop();
        vis[tt] = 0;
        for(int i=0; i<edge[tt].size(); i++)
        {
            int mm = edge[tt][i].v;
            if(dist[mm] > dist[tt] + edge[tt][i].cost)
            {
                dist[mm] = dist[tt] + edge[tt][i].cost;
                if(!vis[mm])
                {
                    vis[mm] = 1;
                    que.push(mm);
                }
            }
        }
    }
    memset(vis,0,sizeof(vis));
    memset(dist1,INF,sizeof(dist1));
    while(!que.empty()) que.pop();
    vis[start] = 1;
    dist1[start] = 0;
    que.push(start);
    while(!que.empty())
    {
        int tt = que.front();
        que.pop();
        vis[tt] = 0;
        for(int i=0; i<edge1[tt].size(); i++)
        {
            int mm = edge1[tt][i].v;
            if(dist1[mm] > dist1[tt] + edge1[tt][i].cost)
            {
                dist1[mm] = dist1[tt] + edge1[tt][i].cost;
                if(!vis[mm])
                {
                    vis[mm] = 1;
                    que.push(mm);
                }
            }
        }
    }
    return ;
}
int main()
{
//#ifndef ONLINE_JUDGE
//          freopen("in.txt","r",stdin);
//#endif // ONLINE_JUDGE
    scanf("%d %d %d",&n,&m,&p);
    int u,v,w;
    for(int i=0; i<m; i++)
    {
        scanf("%d %d %d",&u,&v,&w);
        addedge(u,v,w);
    }
    SPFA(p);
    int mmax = -INF;
    for(int i=1; i<=n; i++)
    {
        if(dist[i] + dist1[i] > mmax && dist[i] != INF && dist1[i]!= INF )
            mmax = dist[i] + dist1[i];
    }
    printf("%d
",mmax);
    return 0;
}
    View Code
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  • 原文地址：https://www.cnblogs.com/chenyang920/p/4411693.html
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