UVA 10720 Graph Construction(贪心 + Havel-Hakimi定理)

Problem C	Graph Construction
Time Limit	2 Seconds

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	Output for Sample Input
4 3 3 3 3 6 2 4 5 5 2 1 5 3 2 3 2 1 0	Possible Not possible Not possible

Problemsetter: Md. Bahlul Haider
Judge Solution: Mohammed Shamsul Alam
Special thanks to Tanveer Ahsan

题意：给定n个点。和每个点的度数，判断这些点能不能连成一个图。

思路：贪心中的Havel-Hakimi定理，详细参考http://sbp810050504.blog.51cto.com/2799422/883904

代码:

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int n, v[10005];
int cmp(int a, int b) {
    return a > b;
}

bool solve() {
    while (v[0] != 0) {
	for (int i = 1; i <= v[0]; i ++) {
	    v[i] --;
	    if (v[i] < 0)
		return false;
	}
	v[0] = 0;
	sort(v, v + n, cmp);
    }
    return true;
}
int main() {
    while (~scanf("%d", &n) && n) {
	memset(v, 0, sizeof(v));
	for (int i = 0; i < n ; i ++)
	    scanf("%d", &v[i]);
	sort(v, v + n, cmp);
	if (solve()) printf("Possible\n");
	else printf("Not possible\n");
    }
    return 0;
}