zoukankan      html  css  js  c++  java
  • [CF1082D]Maximum Diameter Graph

    题目描述 Description###

    Graph constructive problems are back! This time the graph you are asked to build should match the following properties.
    The graph is connected if and only if there exists a path between every pair of vertices.
    The diameter (aka "longest shortest path") of a connected undirected graph is the maximum number of edges in the shortest path between any pair of its vertices.
    The degree of a vertex is the number of edges incident to it.
    Given a sequence of n integers (a_1,a_2,…,a_n) construct a connected undirected graph of n vertices such that:

    the graph contains no self-loops and no multiple edges;
    the degree di of the (i) -th vertex doesn't exceed ai (i.e. (d_i ≤a_i) );
    the diameter of the graph is maximum possible.
    Output the resulting graph or report that no solution exists.

    输入描述 Input Description###

    The first line contains a single integer n (3≤n≤500) — the number of vertices in the graph.

    The second line contains n integers $a_1,a_2,…,a_n $ ((1≤a_i≤n−1) ) — the upper limits to vertex degrees.

    输出描述 Output Description###

    Print "NO" if no graph can be constructed under the given conditions.

    Otherwise print "YES" and the diameter of the resulting graph in the first line.

    The second line should contain a single integer m — the number of edges in the resulting graph.

    The i-th of the next m lines should contain two integers (v_i,u_i) ((1≤v_i,u_i≤n, v_i≠u_i) ) — the description of the i-th edge. The graph should contain no multiple edges — for each pair (x,y) you output, you should output no more pairs (x,y) or (y,x).

    样例输入 Sample Input###

    5
    1 4 1 1 1
    

    样例输出 Sample Output###

    YES 2
    4
    1 2
    3 2
    4 2
    5 2
    

    数据范围及提示 Data Size & Hint###

    见上面

    之前的一些废话###

    明天考高代,所以这个比赛根本没人参加,我也是就把这道题首刀了之后就去复习了

    题解###

    为了使我们的直径最长,我们通过画图发现直径两端的点度数为1,所以我们需要把度数最小的两个点放到直径两端,然后剩下的所有度数大于1的点都可以充当直径上的点,(可以证明这些点不放在直径上都是浪费),然后剩下的那些度数为1的点就用来充当直径上的分岔就好了。
    最后一定是一棵树。

    代码###

    #include<iostream>
    #include<algorithm>
    #include<cmath>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<queue>
    using namespace std;
    #define mem(a,b) memset(a,b,sizeof(a))
    typedef long long LL;
    typedef pair<int,int> PII;
    inline int read()
    {
    	int x=0,f=1;char c=getchar();
    	while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
    	while(isdigit(c)){x=x*10+c-'0';c=getchar();}
    	return x*f;
    }
    bool e[510][510];
    PII a[510];
    int n,first,last,list[510],dm,val[510];
    bool ok;
    void add(int A,int B){e[A][B]=e[B][A]=1;}//printf("add:%d %d
    ",A,B);}
    int main()
    {
    	n=read();
    	for(int i=1;i<=n;i++)a[i].first=read(),a[i].second=i;
    	sort(a+1,a+n+1);
    	first=a[1].second;last=a[2].second;
    	list[++dm]=first;val[dm]=a[1].first;
    	for(int i=3;i<=n;i++)
    		if(a[i].first>=2)list[++dm]=a[i].second,val[dm]=a[i].first,add(list[dm],list[dm-1]);
    	list[++dm]=last;val[dm]=a[2].first;add(list[dm],list[dm-1]);
    	int pos=3;
    	for(int i=2;i<=dm-1;i++)
    		for(int j=1;j<=val[i]-2;j++)
    		{
    			if(a[pos].first==1)add(list[i],a[pos].second),pos++;//printf("pos %d
    ",pos);
    			if(a[pos].first>1)break;
    		}
    	if(a[pos].first==1){
    		printf("NO
    ");
    		return 0;
    	}
    	int cnt=0;
    	printf("YES %d
    ",dm-1);
    	for(int i=1;i<=n;i++)
    		for(int j=i+1;j<=n;j++)if(e[i][j])cnt++;
    	printf("%d
    ",cnt);
    	for(int i=1;i<=n;i++)
    		for(int j=i+1;j<=n;j++)if(e[i][j])printf("%d %d
    ",i,j); 
    	return 0;
    }
    
    

    总结###

    虽然构造题做的少(基本没做过),但是这道题我觉得还是比较简单,多多画图,就知道咋做了。

  • 相关阅读:
    Attach Files to Objects 将文件附加到对象
    Provide Several View Variants for End-Users 为最终用户提供多个视图变体
    Audit Object Changes 审核对象更改
    Toggle the WinForms Ribbon Interface 切换 WinForms 功能区界面
    Change Style of Navigation Items 更改导航项的样式
    Apply Grouping to List View Data 将分组应用于列表视图数据
    Choose the WinForms UI Type 选择 WinForms UI 类型
    Filter List Views 筛选器列表视图
    Make a List View Editable 使列表视图可编辑
    Add a Preview to a List View将预览添加到列表视图
  • 原文地址:https://www.cnblogs.com/FYH-SSGSS/p/11876234.html
Copyright © 2011-2022 走看看