CodeForces 673B Problems for Round（比赛出题规则）

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本文探讨了Codeforces如何根据题目难度及相似性合理分配至不同级别比赛的问题，旨在寻找所有可能的分配方案。

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http://codeforces.com/problemset/problem/673/B

There are n problems prepared for the next Codeforces round. They are arranged in ascending order by their difficulty, and no two problems have the same difficulty. Moreover, there are m pairs of similar problems. Authors want to split problems between two division according to the following rules:

Problemset of each division should be non-empty.
Each problem should be used in exactly one division (yes, it is unusual requirement).
Each problem used in division 1 should be harder than any problem used in division 2.
If two problems are similar, they should be used in different divisions.

Your goal is count the number of ways to split problem between two divisions and satisfy all the rules. Two ways to split problems are considered to be different if there is at least one problem that belongs to division 1 in one of them and to division 2 in the other.

Note, that the relation of similarity is not transitive. That is, if problem i is similar to problem j and problem j is similar to problem k, it doesn't follow that i is similar to k.

Input

The first line of the input contains two integers n and m (2 ≤ n ≤ 100 000, 0 ≤ m ≤ 100 000) — the number of problems prepared for the round and the number of pairs of similar problems, respectively.

Each of the following m lines contains a pair of similar problems ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi). It's guaranteed, that no pair of problems meets twice in the input.

Output

Print one integer — the number of ways to split problems in two divisions.

Note

In the first sample, problems 1 and 2 should be used in division 2, while problems 4 and 5 in division 1. Problem 3 may be used either in division 1 or in division 2.

In the second sample, all pairs of problems are similar and there is no way to split problem between two divisions without breaking any rules.

Third sample reminds you that the similarity relation is not transitive. Problem 3 is similar to both 1 and 2, but 1 is not similar to 2, so they may be used together.

题意：

Codeforces出题有 Div1 和 Div2 等级，D1中最简单的题目也要比 D2中的难（雨神说我现在的水平做不出来Div1的题）。

给定一些相似的题目，问可以出多少套题目，

数字代表难度（越大越难）。

思路：

模拟试一下，参考官方的题解。

Code：

#include<cstdio>
#include<cstring>
const int MYDD=1103;

int MAX(int x,int y) {
	return x>y? x:y;
}

int MIN(int x,int y) {
	return x<y? x:y;
}

int main() {
	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF) {
		int div1=n,div2=1;//div1题目等级高
		int max,min;
		while(m--) {
			int a,b;
			scanf("%d%d",&a,&b);//类似的题目
			min=MIN(a,b);
			max=MAX(a,b);
			div1=MIN(div1,max);//div1题目中最简单的题目
			div2=MAX(div2,min);//div2题目中最难的题目
		}
		if(div1-div2<0) {//难题中最简单的题目没有容易题中最难的难
			puts("0");
		} else {
			printf("%d\n",div1-div2);
		}
	}
	return 0;
}
/*By: Shyazhut*/