UVA 11865 最小树形图+二分
http://vjudge.net/contest/view.action?cid=49353#problem/A
During 2009 and 2010 ICPC world finals, the contest was webcasted via world wide web. Seeing this, some contest organizers from Ajobdesh decided that, they will provide a live stream of their contests to every university in Ajobdesh. The organizers have decided that, they will provide best possible service to them. But there are two problems:
1. There is no existing network between universities. So, they need to build a new network. However, the maximum amount they can spend on building the network is C.
2. Each link in the network has a bandwidth. If, the stream’s bandwidth exceeds any of the link’s available bandwidth, the viewers, connected through that link can’t view the stream.
Due to the protocols used for streaming, a viewer can receive stream from exactly one other user (or the server, where the contest is organized). That is, if you have two 128kbps links, you won’t get 256kbps bandwidth, although, if you have a stream of 128kbps, you can stream to any number of users at that bandwidth.
Given C, you have to maximize the minimum bandwidth to any user.
Input
First line of input contains T(≤50), the number of test cases. This is followed by T test cases.
Each test case starts with an integer N,M,C(1≤N≤60,1≤M≤104,1≤C≤109), the number of universities and the number of possible links, and the budget for setting up the network. Each university is identified by an integer between 0 and N-1, where 0 is the server.
Next M lines each contain 4 integers, u,v,b,c(0≤u, v<N, 1≤b, c≤106, u!=v), describing a possible link from university u to university v, that has the bandwidth of b kbps and of cost c. All links are unidirectional.
There is a blank line before each test case.
Output
For each test case, output the maximum bandwidth to stream. If it’s not possible, output “streaming not possible.”.
Sample Input Output for Sample Input
| 3
3 4 300 0 1 128 100 1 2 256 200 2 1 256 200 0 2 512 300
3 4 500 0 1 128 100 1 2 256 200 2 1 256 200 0 2 512 300
3 4 100 0 1 128 100 1 2 256 200 2 1 256 200 0 2 512 300 |
128 kbps 256 kbps streaming not possible.
|
Problemsetter: Manzurur Rahman Khan, Special Thanks: Jane Alam & Derek Kisman
分析：如果已知最小宽带，则问题转化为：如果唧哝小于此宽带的网线，是否可以搭建成功网络？利用最小树形图，不难做出判断。只需求出从零出发的最小树形图，判断权值和是否超过cost即可。由于最小宽带越小越容易有解，所以只需要二分这个最小带宽。
#include <cstdio>
#include <iostream>
#include <math.h>
#include <cstring>
using namespace std;
//======================================================================
//最小树形图（有向图的最小生成树）朱刘算法模板
const int N=101,M=11000,inf=2147483647;
struct edge
{
int u,v;
int w,c;
}e[M*5];
int n,m,id[N],pre[N],v[N],cost;
int inw[N];
bool zhu_liu(int root,int mid,int n)
{
int s,t,idx=n;
long long ans=0;
while (idx)
{
for (int i=0; i<n; ++i)
inw[i]=inf,id[i]=-1,v[i]=-1;
for (int i=0; i<m; ++i)
{
s=e[i].u;
t=e[i].v;
if (e[i].c>=inw[t] || s==t||e[i].w<mid)
continue;
pre[t]=s;
inw[t]=e[i].c;
}
inw[root]=0;
pre[root]=root;
for (int i=0; i<n; ++i)
{
if (inw[i]==inf)
return false;
ans+=inw[i];
}
idx=0;
for (int i=0; i<n; ++i)
if (v[i]==-1)
{
t=i;
while (v[t]==-1)
v[t]=i,t=pre[t];
if (v[t]!=i || t==root)
continue;
id[t]=idx;
for(s=pre[t]; s!=t; s=pre[s])
id[s]=idx++;
}
if(idx==0)
continue;
for(int i=0; i<n; ++i)
if (id[i]==-1) id[i]=idx++;
for(int i=0; i<m; ++i)
{
e[i].c-=inw[e[i].v];
e[i].u=id[e[i].u];
e[i].v=id[e[i].v];
}
n=idx;
root=id[root];
}
if(ans<=cost)
return true;
else
return false;
}
//==============================================================
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int maxb=0,minb=inf;
scanf("%d%d%d",&n,&m,&cost);
for(int i=0;i<m;i++)
{
scanf("%d%d%d%d",&e[i].u,&e[i].v,&e[i].w,&e[i].c);
e[i+m].u=e[i].u,e[i+m].v=e[i].v,e[i+m].w=e[i].w,e[i+m].c=e[i].c;
maxb=max(maxb,e[i].w);
minb=min(minb,e[i].w);
}
if(zhu_liu(0,minb,n)==false)
printf("streaming not possible.\n");
else
{
int l=minb,r=maxb;
int mid;
while(l<r)
{
mid=l+(r-l+1)/2;
for(int i=0;i<m;i++)
e[i]=e[i+m];
if(zhu_liu(0,mid,n)==true)
l=mid;
else
r=mid-1;
}
printf("%d kbps\n",r);
}
}
return 0;
}