Stockbroker Grapevine

来源：互联网发布：论文网站知乎编辑：程序博客网时间：2024/06/10 01:29

题目链接：http://vjudge.net/contest/view.action?cid=47127#problem/B

题目大意：

输入一个有向带权图，求一个起点，使其能遍历所有顶点并且时间最短。即求出其最小生成图的半径。

思路：

先要用floyd算法求出这个有向图的最小生成图。然后算出各个顶点的ecc（eccentricity），即可知图的radius。

算法步骤：

步骤1：floyd算法对图进行整理。

//floydfor(int k = 1; k <= n; k++)    for(int i = 1; i <= n; i++)        for(int j = 1; j <= n; j++)            if(dis[i][j] > dis[i][k] + dis[k][j])                dis[i][j] = dis[i][k] + dis[k][j];

复杂度为O（n^3）

步骤2：计算各点的ecc。

//computing eccfor(int i = 1; i <= n ; i++){for(int j = 1; j <= n; j++){//cout<<dis[i][j]<<" ";if(dis[i][j] > ecc[i])    ecc[i] = dis[i][j];}//cout<<endl;}

复杂度O(n^2)

步骤3，找出中心点，并判断图是否连通，若连通，则输出该点序号及生成图的半径。

int cen = 1;//center point//choose the one with minimum eccfor(int i = 1 ; i <= n ; i ++){//cout<<"ecc["<<i<<"] = "<<ecc[i]<<endl;if(ecc[i] < ecc[cen])        cen = i;}//outputif( ecc[cen] == INF)//disconnected    cout << "disjoint" << endl;else cout << cen << " " << ecc[cen]<<endl;

完整代码如下：

#include <iostream>#include "memory.h"#define INF 0x3f3f3f3fusing namespace std;int dis[105][105];//distance matrixint ecc[105];//max eccentricity for each people int main(){    int n;    while(cin>>n, n != 0)//n people    {memset(dis,0x3f,sizeof(dis));//initializationmemset(ecc,0,sizeof(ecc));int m;//connecting number for each peoplefor(int i = 1; i <= n; i ++)//n people{cin>>m;int k,value;dis[i][i] = 0;if (m != 0)    for (int j = 1; j <= m; j++)    {    cin>>k>>value;    dis[i][k] = value;    }}//floydfor(int k = 1; k <= n; k++)    for(int i = 1; i <= n; i++)        for(int j = 1; j <= n; j++)            if(dis[i][j] > dis[i][k] + dis[k][j])                dis[i][j] = dis[i][k] + dis[k][j];//computing eccfor(int i = 1; i <= n ; i++){for(int j = 1; j <= n; j++){//cout<<dis[i][j]<<" ";if(dis[i][j] > ecc[i])    ecc[i] = dis[i][j];}//cout<<endl;}int cen = 1;//center point//choose the one with minimum eccfor(int i = 1 ; i <= n ; i ++){//cout<<"ecc["<<i<<"] = "<<ecc[i]<<endl;if(ecc[i] < ecc[cen])        cen = i;}//outputif( ecc[cen] == INF)//disconnected    cout << "disjoint" << endl;else cout << cen << " " << ecc[cen]<<endl;}    return 0;}

评测系统上运行结果：Accepted，运行时间0ms，占用内存260KB.

0 0