hdu1839 二分+最短路

Delay Constrained Maximum Capacity Path

Time Limit: 10000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 373    Accepted Submission(s): 149

Problem Description

Consider an undirected graph with N vertices, numbered from 1 to N, and M edges. The vertex numbered with 1 corresponds to a mine from where some precious minerals are extracted. The vertex numbered with N corresponds to a minerals processing factory. Each edge has an associated travel time (in time units) and capacity (in units of minerals). It has been decided that the minerals which are extracted from the mine will be delivered to the factory using a single path. This path should have the highest capacity possible, in order to be able to transport simultaneously as many units of minerals as possible. The capacity of a path is equal to the smallest capacity of any of its edges. However, the minerals are very sensitive and, once extracted from the mine, they will start decomposing after T time units, unless they reach the factory within this time interval. Therefore, the total travel time of the chosen path (the sum of the travel times of its edges) should be less or equal to T.

 

Input

The first line of input contains an integer number X, representing the number of test cases to follow. The first line of each test case contains 3 integer numbers, separated by blanks: N (2 <= N <= 10.000), M (1 <= M <= 50.000) and T (1 <= T <= 500.000). Each of the next M lines will contain four integer numbers each, separated by blanks: A, B, C and D, meaning that there is an edge between vertices A and B, having capacity C (1 <= C <= 2.000.000.000) and the travel time D (1 <= D <= 50.000). A and B are different integers between 1 and N. There will exist at most one edge between any two vertices.

 

Output

For each of the X test cases, in the order given in the input, print one line containing the highest capacity of a path from the mine to the factory, considering the travel time constraint. There will always exist at least one path between the mine and the factory obbeying the travel time constraint.

 

Sample Input

  
  

   
   2
2 1 10
1 2 13 10
4 4 20
1 2 1000 15
2 4 999 6
1 3 100 15
3 4 99 4
  
  

 

Sample Output

  
  

   
   13
99
  
  

 

Author

Mugurel Ionut Andreica

 

Source

Politehnica University of Bucharest Local Team Contest 2007

 

Recommend

lcy

 

容量越大时间越长~以此二分~感觉不是很严谨额~~

先是未优化dijkstra10S还是超时~~~换spfa过~~~没有试优先队列优化的dijkstra

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#define inf 99999999
using namespace std;
struct node
{
    int u,v,w,c;
};
node e[100005];
int first[10005];
int next[100005];
bool inq[10005];
int d[10005],n;
void add_edge(int u,int v,int c,int w,int i)
{
    e[i].u=u;
    e[i].v=v;
    e[i].w=w;
    e[i].c=c;
    next[i]=first[u];
    first[u]=i;
}
void spfa(int ff)
{
    memset(inq,0,sizeof(inq));
    int i;
    for(i=0;i<=n;i++)
        d[i]=inf;
    d[1]=0;
    queue<int> q;
    q.push(1);
    while(!q.empty())
    {
        int x=q.front(),y;
        q.pop();
        inq[x]=false;
        for(y=first[x];y!=-1;y=next[y])
        {
            int u=e[y].u;
            int v=e[y].v;
            if(d[v]>=d[x]+e[y].w&&e[y].c>=ff)
            {
                d[v]=d[x]+e[y].w;
                if(!inq[v])
                {
                    inq[v]=true;
                    q.push(v);
                }
            }
        }
    }
}
int main()
{
    int x,cas=0;
    scanf("%d",&x);
    for(cas=0;cas<x;cas++)
    {
        int m,t;
        scanf("%d%d%d",&n,&m,&t);
        int i;
        memset(first,-1,sizeof(first));
        memset(next,-1,sizeof(next));
        int Min=inf,Max=0;
        int k=0;
        for(i=0;i<m;i++)
        {
            int u,v,w,c;
            scanf("%d%d%d%d",&u,&v,&c,&w);
            if(c>Max)
                Max=c;
            if(c<Min)
                Min=c;
            add_edge(u,v,c,w,k++);
            add_edge(v,u,c,w,k++);
        }
        int ans;
        while(Min<=Max)
        {
            int l=(Min+Max)/2;
            spfa(l);
            if(d[n]<=t)
            {
                Min=l+1;
                ans=l;
            }
            else
                Max=l-1;
        }
        printf("%d\n",ans);
    }
    return 0;
}