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zoj 2281 Way to Freedom

ZOJ Problem Set - 2281
Way to Freedom

Time Limit: 2 Seconds      Memory Limit: 32768 KB

Background

"Now, you are getting closer and closer to Your Gate to Freedom!

And you will pass the most dangerous paths! Be careful!"

We're in a maze now. There're many rooms floating in the sky connected with wooden bridges. Bridges seems unstable...


Problem

At first, I'm in room X, and I want to go to room Y. There're N (N <= 100,000) rooms connected with M (M <= 1,000,000) edges. Each edge has a limit of weight, if weight exceeds this value, the edge will split, and drop me into the river.

How many things I can take to room Y at most?


Input

This problem contains multiple test cases.

Each test case begins with two integers N and M, then M lines, each line has 3 numbers: X, Y and Weight (<=2,000,000,000).


Output

For each test case, output the maximum weight I can take.


Sample Input

6 6
1 5 2000
2 4 5000
2 5 3300
3 4 2400
3 6 2200
4 6 6000
3 5


Sample Output

2400


Author: JIANG, Yanyan
Contest: A Great Beloved and My Gate to Freedom
There is a cx, there is a love_cx.


Source: JIANG, Yanyan's Contest #2
Submit    Status
//1838048 2009-04-19 12:27:25 Accepted  2281 C++ 580 6588 Wpl 
//更牛的Dijkstra+优先队列解决最大网络流问题
#include <iostream>
#include 
<queue>
#include 
<vector>
#define MAX 100005
using namespace std;
typedef 
struct node
{
    
int heap;
    
int dis;
    node()
    {}
    node(
int h,int d)
    {
        heap
=h;
        dis
=d;
    }
    friend 
bool operator <(node a,node b)  //用的是大堆
    {
        
return a.dis<b.dis;
    }
}Point;
priority_queue
<Point>Q;
Point temp;
vector
<Point>G[MAX];
int n,m,start,end,dis[MAX],zz=1;
void Init()
{
    
int i,a,b,s;
    
for(i=1;i<=n;i++)  //初始化
        G[i].clear();
    
while(m--)
    {
        scanf(
"%d%d%d",&a,&b,&s);
        G[a].push_back(node(b,s));
        G[b].push_back(node(a,s));
    }
    scanf(
"%d%d",&start,&end);
}
int GetMin(int a,int b)
{
    
if(a==-1)
        
return b;
    
else
    {
        
if(a>b)
            
return b;
        
else
            
return a;
    }
}
void Dijkstra()
{
    
int i,tp,td,k;
    
for(i=1;i<=n;i++)
    {
        dis[i]
=-1;
    }
    
while(!Q.empty())
        Q.pop();
    Q.push(node(start,
-1));
    
while(!Q.empty())
    {
        temp
=Q.top();
        k
=temp.heap;
        Q.pop();
        
if(k==end)
        {
            printf(
"%d\n",temp.dis);
            
return ;
        }
        
for(i=0;i<G[temp.heap].size();i++)
        {
            tp
=G[temp.heap][i].heap;
            td
=G[temp.heap][i].dis;
            
if(GetMin(dis[k],td)>dis[tp])
            {
                dis[tp]
=GetMin(dis[k],td);
                Q.push(node(tp,dis[tp]));
            }
        }
    }
    printf(
"0\n");
}
int main()
{
    
while(scanf("%d%d",&n,&m)!=EOF)
    {
        Init();
        Dijkstra();
    }
    
return 0;
}
posted @ 2009-05-21 09:48  往往  阅读(332)  评论(0编辑  收藏  举报