POJ Stockbroker Grapevine(最短路算法)

                                                                        Stockbroker Grapevine

题目链接：Click Here~

题目分析：

   不知道这题是为了考英语阅读理解还是考算法。国外的题目做起来就是蛋疼。

   题目意思是：给你n个点，叫你求出从n个点中选出一个点到到其他n-1个点的最长时间time，而又从这n个点中选出time最小的。靠！看了好久都没懂那题目啥意思。

算法分析：

   其实本质就是n次最短路的问题。过程被题目描述的高端了。所以，我用了SPFA和FLOYD算法试了一下，都可以过。而且，，，。数据中没有找不到的情况。

Spfa版：

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

struct Edge{
     int from,to,cost;          //从from到to的时间
};
class StockBroker{
   public:
      static const int maxn = 100+5;
      static const int INF = 1e8;
      StockBroker(int n){
          this->n = n;
          for(int i = 0;i <= n;++i)
            G[i].clear();
          edges.clear();
      }
      void Init();
      void AddEdge(int from,int to,int cost);
      int Spfa(int s);
      void Print();
   private:
      int n,m;
      vector<Edge> edges;
      vector<int> G[maxn];
};

void StockBroker::AddEdge(int from,int to,int cost)   // 加边
{
    edges.push_back((Edge){from,to,cost});
    m = edges.size();
    G[from].push_back(m-1);
}
int StockBroker::Spfa(int s)
{
    queue<int> Q;
    int d[maxn];
    bool inq[maxn];
    for(int i = 0;i <= n;++i)
       d[i] = INF,inq[i] = false;
    d[s] = 0;
    Q.push(s);
    while(!Q.empty()){
        int u = Q.front();
        Q.pop();
        inq[u] = false;
        for(int i = 0;i < (int)G[u].size();++i){
           Edge& e = edges[G[u][i]];
           if(d[e.to]>d[u]+e.cost){
              d[e.to] = d[u] + e.cost;
              if(!inq[e.to]){
                  inq[e.to] = true;
                  Q.push(e.to);
               }
           }
        }
    }
    int maxv = -INF;
    for(int i = 1;i <= n;++i)
       maxv = max(maxv,d[i]);
    return maxv;
}
int main()
{
    int n,m;
    while(scanf("%d",&n),n)
    {
        int u,c;
        StockBroker spfa(n);
        for(int i = 1;i <= n;++i){
           scanf("%d",&m);
           while(m--){
              scanf("%d%d",&u,&c);
              spfa.AddEdge(i,u,c);
           }
        }
        int pos=1,ans = 100000; 
        for(int i = 1;i <= n;++i){         //调用n次最短路算法
             int tmp = spfa.Spfa(i);
             if(tmp < ans){
                 pos = i;
                 ans = tmp;
             }
        }
        printf("%d %d\n",pos,ans);
    }
    return 0;
}

Floyd版：

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
using namespace std;

class StockBroker{
    public:
       static const int maxn = 100+5;
       static const int INF = 1e8;
       StockBroker(int n){
           this->n = n;
           for(int i = 0;i <= n;++i)
              for(int j = 0;j <= n;++j)
                 d[i][j] = (i==j?0:INF);
       }
       void Init(int from,int to,int cost);
       void AddEdge(int from,int to,int cost);
       void Floyd();
       void Print();
    private:
       int n,m;
       int d[maxn][maxn];
};
inline void StockBroker::Init(int from,int to,int cost)
{
    d[from][to] = cost;
}
void StockBroker::Floyd()
{
    for(int k = 1;k <= n;++k)
      for(int i = 1;i <= n;++i)
        for(int j = 1;j <= n;++j)
           if(d[i][j] > d[i][k]+d[k][j])
              d[i][j] = d[i][k]+d[k][j];
}
void StockBroker::Print()
{
    int pos=1,ans = INF;
    for(int i = 1;i <= n;++i){
        int maxv = -INF;
        for(int j = 1;j <= n;++j){
          if(maxv < d[i][j])
              maxv = d[i][j];
        }
        if(ans > maxv){
            ans = maxv;
            pos = i;
        }
    }
    printf("%d %d\n",pos,ans);
}
int main()
{
    int n,m;
    while(scanf("%d",&n),n)
    {
        int u,cost;
        StockBroker floyd(n);
        for(int i = 1;i <= n;++i){
           scanf("%d",&m);
           while(m--){
              scanf("%d%d",&u,&cost);
              floyd.Init(i,u,cost);
           }
        }
        floyd.Floyd();
        floyd.Print();
    }
    return 0;
}

POJ Stockbroker Grapevine(最短路算法),布布扣,bubuko.com

POJ Stockbroker Grapevine(最短路算法)

原文：http://blog.csdn.net/zhongshijunacm/article/details/21599611