/*1.对输入做出一定的处理,变成邻接矩阵,是有向图;
2.假设1为源点,求出来1到所有点的最大值,然后以2为源点,求出来到所有的最大值,最后取其最小的值作为
答案;
3.要检查一下是否有不可达的点
这个是maxmin,也就是最大值的最小值
这个题用floyd会很方便但是性能上可能不如dijkstra*/
#include <iostream>
#include <vector>
#include <queue>
#include <utility>
#define N 101
#define Inifite 0x3fffffff
#define pli pair<int,int>//length and index
using namespace std;
int edge[N][N],dist[N],flag[N];
int numOfBrokers;
void initial()
{
for (int i=1;i<=numOfBrokers;i++)
{
for (int j=1;j<=numOfBrokers;j++)
{
if(i==j)
edge[i][j]=0;
else edge[i][j]=Inifite;
}
}
}
void initialDijkstra(int start)
{
for (int i=1;i<=numOfBrokers;++i)
dist[i]=Inifite;
memset(flag,0,sizeof(flag));
dist[start]=0;
flag[start]=1;
}
int findIndexOfMax()
{
int maxIndex=1;
for (int i=2;i<=numOfBrokers;i++)
{
if (dist[i]>dist[maxIndex])
maxIndex=i;
}
return maxIndex;
}
int dijkstra(int start)
{
initialDijkstra(start);
priority_queue<pli,vector<pli>,greater<pli> > q;
q.push(make_pair(0,start));
while (!q.empty())
{
pli temp=q.top();
q.pop();
int minIndex=temp.second;
int minlength=temp.first;
flag[minIndex]=1;
for (int i=1;i<=numOfBrokers;++i)
{
if(!flag[i]&&dist[i]>dist[minIndex]+edge[minIndex][i])
{
dist[i]=dist[minIndex]+edge[minIndex][i];
q.push(make_pair(dist[i],i));
}
}
}
return dist[findIndexOfMax()];
}
int main()
{
//freopen("in.txt","r",stdin);
int from,to,costMinutes,j=1;
while (cin >> numOfBrokers &&numOfBrokers)
{
initial();
//situation just like this:has many times inputs,remember initial some
//necessary variations at first,especially counters or outputs
int minLength=Inifite,minIndex=1,temp,j=1;//every time j will be set as 1;
while (j<=numOfBrokers)
{
int contracts,i=1;
cin >> contracts;
while (i<=contracts)
{
from=j;
cin >> to >> costMinutes;
edge[from][to]=costMinutes;
++i;
}
++j;
}
for (j=1;j<=numOfBrokers;++j)
{
temp=dijkstra(j);
if(temp<minLength)
{
minLength=temp;
minIndex=j;
}
}
if(minLength<Inifite)
cout << minIndex << " " << minLength << endl;
else cout << "disjoint" << endl;
}
}
poj1125之dijksrta
最新推荐文章于 2025-02-24 22:09:39 发布
