HDU - 1839 Delay Constrained Maximum Capacity Path（二分+dijkstra）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1839点击打开链接

Delay Constrained Maximum Capacity Path

Time Limit: 10000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)

Total Submission(s): 2157    Accepted Submission(s): 731


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 lin
cd83
e 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

题目要求求规定时间内从1到n的经过最小路容量的最大值

很明显这种类型的就是二分了

而这种套路一般都是对结果二分

也就是对容量二分

对于djkstra 在每次判断的时候加上对二分条件的判断限制 就可以根据结果判定此次二分的条件能不能找到一条符合最小容量大于当前mid的路径 

满足单调性

因为写成ans=l找了好久错误。。

#include <bits/stdc++.h>
using namespace std;
long long int n,m,ttime;
struct xjy
{
long long int num;
long long int time;
long long int val;
bool operator < (const xjy &r)const
{
return time>r.time;
}
};
vector<xjy > mmap[11111];
long long int djkstra(long long int limit)
{
long long int book[11111];
long long int dis[11111];
for(long long int i=0;i<11111;i++)
{
book[i]=0;
dis[i]=99999999;
}
dis[1]=0;
xjy mid;
mid.num=1;
mid.val=0;
mid.time=0;
priority_queue<xjy > q;
q.push(mid);
while(!q.empty())
{
mid=q.top();
q.pop();
if(book[mid.num])
continue;
book[mid.num]=1;
for(long long int i=0;i<mmap[mid.num].size();i++)
{
long long int now=mid.num;
if(dis[mmap[now][i].num]>dis[now]+mmap[now][i].time&&limit<=mmap[now][i].val)
{
dis[mmap[now][i].num]=dis[now]+mmap[now][i].time;
xjy midmid;
midmid.num=mmap[now][i].num;
midmid.time=dis[mmap[now][i].num];
q.push(midmid);
}
}
}
return dis
<=ttime;
}
int main()
{
long long int t;
scanf("%lld",&t);
while(t--)
{
for(long long int i=0;i<11111;i++)
{
mmap[i].clear();
}
long long int l=99999999,r=0;
scanf("%lld%lld%lld",&n,&m,&ttime);
for(long long int i=0;i<m;i++)
{
long long int mid1,mid2,mid3,mid4;
scanf("%lld%lld%lld%lld",&mid1,&mid2,&mid3,&mid4);
xjy mid;
mid.num=mid2;
mid.time=mid4;
mid.val=mid3;
mmap[mid1].push_back(mid);
mid.num=mid1;
mmap[mid2].push_back(mid);
r=max(r,mid.val);
l=min(l,mid.val);
}
long long int ans=0;
while(l<=r)
{
long long int mid=(l+r)>>1;
if(djkstra(mid))
{
ans=mid;
l=mid+1;
}
else
r=mid-1;
}
printf("%lld\n", ans);
}
}