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。某个子集中，点的权值是这样算的，在该子集中这个点的度除以该点在图中的度。乍看上去似乎无从下手。可以显然知道的是，每个点在图中的权值是很容易算出来的，那我们尝试从图中进行删点，使得当前图的最小权值点的权值变大，显然可以知道要删除最小权值点，为什么呢？因为若删除次小权值点，若次小权值点跟最小权值点有连边，那最小权值点还是新图的最小权值点。若没有连边，那新图的最小权值点也依旧未变。所以只有删除最小权值点才有可能改变新图的最小权值点，也只有这样能让新图的最小权值发生改变。那么到这里就十分明显了，只要每次删除当前图的最小权值点，那么必然可以枚举出一个新图，这个新图的点构成的点集正是我们要的答案。于是这个题就可以做了，我是直接做了两次删除，，第一次得出最大最小权值是多少，第二次枚举到一个新图的最小权值等于最大最小权值，那么很显然这个新图的子集就是答案。
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<<" ";}
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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 arencities, labeled from 1 ton, andmbidirectional roads between them. Currently, there are Life Fibers in every city. In addition, there arekcities 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 citiesSwith no fortresses of Life Fibers. The strength of a cityis defined as (number of neighbors ofxinS) / (total number of neighbors ofx). Here, two cities are called neighbors if they are connnected with a road. The goal is to maximize the strength of the weakest city inS.
Given a description of the graph, and the cities with fortresses, find a non-empty subset that maximizes the strength of the weakest city.
Input
The first line of input contains three integersn,m,k(2≤n≤100000,1≤m≤100000,1≤k≤n-1).
The second line of input containskintegers, representing the cities with fortresses. These cities will all be distinct.
这种精神使人能在旅行中和大自然更加接近，
