UVa10720 （ Graph Construction）（判断是否可化为简单图）

Graph is a collection of edges E and vertices V. 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 are n vertices then n integers 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 by n number of integers. But the reverse is not always true. If you
are given n integers, you have to find out whether this n numbers can represent the degrees of n vertices
of a graph.
Input
Each line will start with the number n (≤ 10000). The next n integers will represent the degrees of n
vertices of the graph. A ‘0’ input for n will indicate end of input which should not be processed.
Output
If the n integers can represent a graph then print ‘Possible’. Otherwise print ‘Not possible’. Output
for each test case should be on separate line.
Sample Input
4 3 3 3 3
6 2 4 5 5 2 1
5 3 2 3 2 1
0
Sample Output
Possible
Not possible

Not possible

这道题和杭电2454几乎一模一样。

#include<stdio.h>
#include<algorithm>
using namespace std;
int a[10010];
int cmp(int x,int y)
{
	return x>y;
}
int main()
{
	int n,sum,flag;
	while(scanf("%d",&n),n)
	{
		sum=flag=0;
		for(int i=0;i<n;i++)
		{
		    scanf("%d",&a[i]);
		    sum+=a[i];
		}
		sort(a,a+n,cmp);
		if(sum&1||a[0]>=n)
		{
			printf("Not possible\n");
			continue;
		}
		for(int i=0;i<n-1;i++)
		{
			sort(a+i,a+n,cmp);
			for(int j=i+1;j<=i+a[i];j++)
			{
				a[j]--;
				if(a[j]<0)
				{
					flag=1;
					break;
				}
			}
			if(flag)
			    break;
		}
		if(a[n-1]!=0)
		    flag=1;
		printf(flag?"Not possible\n":"Possible\n");
		
	}
}