Degree Sequence of Graph G

    Problem DescriptionWang Haiyang is a strong and optimistic Chinese youngster. Although born and brought up in the northern inland city Harbin, he has deep love and yearns for the boundless oceans. After graduation, he came to a coastal city and got a job in a marine transportation company. There, he held a position as a navigator in a freighter and began his new life.

    The cargo vessel, Wang Haiyang worked on, sails among 6 ports between which exist 9 routes. At the first sight of his navigation chart, the 6 ports and 9 routes on it reminded him of Graph Theory that he studied in class at university. In the way that Leonhard Euler solved The Seven Bridges of Knoigsberg, Wang Haiyang regarded the navigation chart as a graph of Graph Theory. He considered the 6 ports as 6 nodes and 9 routes as 9 edges of the graph. The graph is illustrated as below.

    AccZ喎?http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">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"s put forward such a definition: provided that a non-negative integer sequence is the degree sequence of a graph without any parallel edge or ring, that is, a simple graph, the sequence is draw-possible, otherwise, non-draw-possible. Now the problem faced with Wang Haiyang is how to test whether a non-negative integer sequence is draw-possible or not. Since Wang Haiyang hasn‘t studied Algorithm Design course, it is difficult for him to solve such a problem. Can you help him?

    InputThe first line of input contains an integer T, indicates the number of test cases. In each case, there are n+1 numbers; first is an integer n (n<1000), which indicates there are n integers in the sequence; then follow n integers, which indicate the numbers of the degree sequence.

    OutputFor each case, the answer should be "yes"or "no" indicating this case is "draw-possible" or "non-draw-possible"

    Sample Input

26 4 4 3 3 2 24 2 1 1 1

    Sample Output

yesno

    Source2008 Asia Regional Harbin

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#include<stdio.h>#include<string.h>#include<stdlib.h>int cmp(const void *a,const void *b){	return *(int *)b-*(int *)a;}int a[10010];int main(){	int t;	scanf("%d",&t);	while(t--)	{		int n,i,j,sum=0;		scanf("%d",&n);		for(i=0;i<n;i++)		{			scanf("%d",&a[i]);			sum+=a[i];		}		if(sum%2)		{			printf("no\n");			continue;		}		for(i=0;i<n;i++)		{			if(a[i]>=n)				break;		}		if(i<n)		{			printf("no\n");			continue;		}		int flag=0;		for(i=0;i<n;i++)		{			int cnt=0;			qsort(a,n,sizeof(a[0]),cmp);			for(j=1;j<n;j++)			{				if(a[0]==cnt)					break;				a[j]--;				cnt++;				if(a[j]<0)				{					flag=1;					break;				}			}			if(flag)				break;			if(cnt==0)				continue;			a[0]-=cnt;		}		if(flag)		{			printf("no\n");			continue;		}		for(i=0;i<n;i++)		{			if(a[i])				break;		}		if(i<n)			printf("no\n");		else			printf("yes\n");	}}