2013亚洲区域赛长沙站 ZOJ 3732 Graph Reconstruction

题目链接

Graph Reconstruction

题意

给你无向图每个点的度数, 问是否存在唯一解, 存在输出唯一解, 多解输出两个, 无解输出IMPOSSIBLE

思路

这里用到了 Havel-Hakimi定理, 实际上就是按照度数构图的一种贪心策略. 这样能判断出来一个解或无解, 多解的情况, 只要在比较构图时最后面两个点的当前度数一样, 那么选其一都是可行的, 但是选其一的情况有保证了另外一个肯定没有连上, 所以能找到其它解.

代码

#include <bits/stdc++.h>

#define LL long long
#define INF 0x3f3f3f3f
#define eps 1e-8

using namespace std;

struct Node{
	int val, pos;
}d[105], b[105];
pair<int, int> res1[10005], res2[10005];
int cnt1, cnt2;
bool compare(Node x, Node y){
	return x.val > y.val;
}

int main(){
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
	int n;
	while (~scanf("%d", &n)){
		bool noans = false;
		cnt1 = cnt2 = 0;
		for (int i = 1; i <= n; i++){
			scanf("%d", &d[i].val);
			d[i].pos = i;
			if (d[i].val > n - 1) noans = true;
			b[i] = d[i];
		}
		sort(d + 1, d + n + 1, compare);
		bool isunique = true;
		while (!noans){
			if (d[1].val == 0) break;
			for (int i = 2; i < 2 + d[1].val; i++){
				d[i].val--;
				res1[cnt1++] = make_pair(d[1].pos, d[i].pos);
				if (d[i].val < 0){
					noans = true;
					break;
				}
			}
			if (d[2 + d[1].val].val - d[1 + d[1].val].val == 1){
				isunique = false;
			}
			d[1].val = 0;
			sort(d + 1, d + n + 1, compare);
		}
		if (noans){
			printf("IMPOSSIBLE\n");
			continue;
		}
		if (isunique){
			printf("UNIQUE\n");
		}
		else{
			printf("MULTIPLE\n");
			for (int i = 1; i <= n; i++){
				d[i] = b[i];
			}
			sort(d + 1, d + n + 1, compare);
			bool first = true;
			while (d[1].val != 0){
				for (int i = 2; i < 1 + d[1].val; i++){
					d[i].val--;
					res2[cnt2++] = make_pair(d[1].pos, d[i].pos);
				}
				if (first && d[1 + d[1].val].val == d[2 + d[1].val].val){
					first = false;
					d[2 + d[1].val].val--;
					res2[cnt2++] = make_pair(d[1].pos, d[2 + d[1].val].pos);
				}
				else{
					d[1 + d[1].val].val--;
					res2[cnt2++] = make_pair(d[1].pos, d[1 + d[1].val].pos);
				}
				d[1].val = 0;
				sort(d + 1, d + n + 1, compare);
			}
		}
		printf("%d %d\n", n, cnt1);
		for (int i = 0; i < cnt1; i++){
			printf("%d ", res1[i].first);
		}
		printf("\n");
		for (int i = 0; i < cnt1; i++){
			printf("%d ", res1[i].second);
		}
		printf("\n");
		if (!isunique){
			printf("%d %d\n", n, cnt2);
			for (int i = 0; i < cnt2; i++){
				printf("%d ", res2[i].first);
			}
			printf("\n");
			for (int i = 0; i < cnt2; i++){
				printf("%d ", res2[i].second);
			}
			printf("\n");
		}
	}
}