【Codeforces Round #351 Div. 2】 673B Problems for Round

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.

【题目分析】

把一列数划分成两个集合，只能从中间分开，要求左面的比右面的要小，显然，可以维护一个左面最大值和右面最小值，看看是否冲突就好了。

【代码】

#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
using namespace std;
int main()
{
int n,m;
scanf("%d%d",&n,&m);
int lmax=1,rmin=n;
for (int i=1;i<=m;++i){
int u,v;
scanf("%d%d",&u,&v);
lmax=max(min(u,v),lmax);
rmin=min(max(u,v),rmin);
}
if (lmax>=rmin) {
printf("0\n"); return 0;
}
else {
printf("%d\n",rmin-lmax); return 0;
}
}