POJ 2808 校门外的树

利用0和1模拟有树和无树的状态，应该是比较舒服的一种方法。利用排序后区间之间的关系做显得有点没必要

#include<stdio.h>
#include<string.h>

int main(void)
{
  int L, M, tree[10005], a, b, count;

  while (scanf("%d %d", &L, &M) != EOF) {
    count = 0;
    memset(tree, 0, sizeof(tree));
    for (int i = 0; i < M; i++) {
      scanf("%d %d", &a, &b);

      for (int i = a; i <= b; i++) {
        tree[i] = 1;
      }
    }

    for (int i = 0; i <= L; i++) {
      if (tree[i] == 0) {
        count++;
      }
    }

    printf("%d\n", count);
  }
}