标签：
题意大概是，给出一个图，保证每个点至少有一条边以及任意两点间最多一条边。很显然这个图有众多点集，若我们给每个点定义一个权值，那每个点集都有一个最小权值点，现在要求出一个点集，这个点集的最小权值点尽可能的大。维护一个小堆就好了，时间复杂度是2(n+m)log(n+m)
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
struct node
{
int no;
double val;
node(){}
node(int no,double val)
{
this->no=no;
this->val=val;
}
bool operator <(node one)const
{
return val>one.val;
}
};
priority_queue<node>q[2];
bool fb[100010],dl[100010];
vector<int>edge[100010];
int d[100010][2],dd[100010];
int main()
{
int n,m,k;
cin>>n>>m>>k;
int sum=k;
while(k--)
{
int t;
cin>>t;
fb[t]=1;
}
while(m--)
{
int a,b;
cin>>a>>b;
d[a][0]++;
d[b][0]++;
if(!fb[a])
d[b][1]++;
if(!fb[b])
d[a][1]++;
edge[a].push_back(b);
edge[b].push_back(a);
}
double mn=1e99;
for(int i=1;i<=n;i++)
if(!fb[i])
{
q[0].push(node(i,double(d[i][1])/d[i][0]));
mn=min(mn,double(d[i][1])/d[i][0]);
}
q[1]=q[0];
int flag=0;
for(int i=0;i<2;i++)
{
memset(dl,0,sizeof(dl));
memset(dd,0,sizeof(dd));
if(i==1&&flag==0)
break;
while(q[i].size())
{
node t=q[i].top();
q[i].pop();
if(dl[t.no])
continue;
if(i==1)
{
if(t.no==flag)
break;
sum++;
}
dl[t.no]=1;
if(i==0&&t.val>mn)
{
mn=t.val;
flag=t.no;
}
for(int j=0;j<edge[t.no].size();j++)
{
int v=edge[t.no][j];
if(dl[v]||fb[v])
continue;
dd[v]++;
q[i].push(node(v,double(d[v][1]-dd[v])/d[v][0]));
}
}
}
cout<<n-sum<<endl;
for(int i=1;i<=n;i++)
if(!fb[i]&&!dl[i])
cout<<i<<" ";
}Nudist Beach is planning a military operation to attack the Life Fibers. In this operation, they will attack and capture several cities which are currently under the control of the Life Fibers.
There are n cities, labeled from 1 to n, and m bidirectional roads between them. Currently, there are Life Fibers in every city. In addition, there are k cities that are fortresses of the Life Fibers that cannot be captured under any circumstances. So, the Nudist Beach can capture an arbitrary non-empty subset of cities with no fortresses.
After the operation, Nudist Beach will have to defend the captured cities from counterattack. If they capture a city and it is connected to many Life Fiber controlled cities, it will be easily defeated. So, Nudist Beach would like to capture a set of cities such that for each captured city the ratio of Nudist Beach controlled neighbors among all neighbors of that city is as high as possible.
More formally, they would like to capture a non-empty set of cities S with no fortresses of Life Fibers. The strength of a city 
is
defined as (number of neighbors of x in S)
/ (total number of neighbors of x). Here, two cities are called neighbors if they are connnected with a road. The goal is to maximize
the strength of the weakest city in S.
Given a description of the graph, and the cities with fortresses, find a non-empty subset that maximizes the strength of the weakest city.
The first line of input contains three integers n,?m,?k (2??≤??n??≤?100?000, 1?≤?m?≤?100?000, 1?≤?k?≤?n?-?1).
The second line of input contains k integers, representing the cities with fortresses. These cities will all be distinct.
The next m lines contain the roads. The i-th of these lines will have 2 integers ai,?bi (1?≤?ai,?bi?≤?n, ai?≠?bi). Every city will have at least one road adjacent to it.
There is no more than one road between each pair of the cities.
The first line should contain an integer r, denoting the size of an optimum set (1?≤?r?≤?n?-?k).
The second line should contain r integers, denoting the cities in the set. Cities may follow in an arbitrary order. This line should not contain any of the cities with fortresses.
If there are multiple possible answers, print any of them.
9 8 4 3 9 6 8 1 2 1 3 1 4 1 5 2 6 2 7 2 8 2 9
3 1 4 5
10 8 2 2 9 1 3 2 9 4 5 5 6 6 7 7 8 8 10 10 4
8 1 5 4 8 10 6 3 7
The first example case achieves a strength of 1/2. No other subset is strictly better.
The second example case achieves a strength of 1. Note that the subset doesn‘t necessarily have to be connected.
版权声明：本文为博主原创文章，未经博主允许不得转载。
标签：
原文地址：http://blog.csdn.net/stl112514/article/details/46965697
