UVa 10720 - Graph Construction

链接：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=113&page=show_problem&problem=1661

原题：

Graph is a collection of edgesEand verticesV. Graph has a wide variety of applications in computer. There are different ways to represent graph in computer. It can be represented by adjacency matrix or by adjacency list. There are some other ways to represent graph. One of them is to write the degrees (the numbers of edges that a vertex has) of each vertex. If there arenvertices thennintegers can represent that graph. In this problem we are talking about simple graph which does not have same endpoints for more than one edge, and also does not have edges with the same endpoint.

Any graph can be represented bynnumber of integers. But the reverse is not always true. If you are givennintegers, you have to find out whether thisnnumbers can represent the degrees ofnvertices of a graph.

Input
Each line will start with the numbern (≤ 10000). The nextnintegers will represent the degrees ofnvertices of the graph. A 0 input for n will indicate end of input which should not be processed.

Output
If thenintegers can represent a graph then print “Possible”. Otherwise print “Not possible”. Output for each test case should be on separate line.

样例输入：

4 3333
6 2 4 552 1
5 3 2 3 2 1
0

样例输出：

Possible
Not possible
Not possible

题目大意：

给出一张有n个点的图，然后再给出每个点的度数。求这个图是否可能存在。

分析与总结：

也是赤裸裸的Havel-Hakimi定理的应用。

代码：

/*
 * UVa: 10720 - Graph Construction
 * 贪心，Havel-Hakimi定理
 * Time: 0.040s
 * Author: D_Double
 *
 */
#include<iostream>
#include<algorithm>
#include<cstdio>
#define MAXN 10005
using namespace std;

int arr[MAXN],n;

bool Havel_Hakimi(){
    for(int i=0; i<n-1; ++i){
        sort(arr+i,arr+n,greater<int>());
        if(i+arr[i] >= n) return false;
        for(int j=i+1; j<=i+arr[i]; ++j){
            --arr[j];
            if(arr[j] < 0) return false;
        }
    }
    if(arr[n-1]!=0) return false;
    return true;
}

int main(){
    while(scanf("%d",&n)&&n){
        int sum=0;
        bool flag=true;
        for(int i=0; i<n; ++i){
            scanf("%d",&arr[i]);
            if(arr[i]>n-1) flag=false;
            sum += arr[i];
        }
        if(sum%2!=0) flag=false;
        if(sum==0 || flag&&Havel_Hakimi()) printf("Possible\n");
        else printf("Not possible\n");
    }
    return 0;
}

 
—— 生命的意义，在于赋予它意义。

 

   
 

 

   原创http://blog.csdn.net/shuangde800，By D_Double (转载请标明)