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
//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;
}