IDIOPEN 2013 D Negative People in Da House

题目地址

题目大意：给出n种进出（p1，p2）情况，一定是先进后出，求房间里最少有多少个人

解题思路：简单模拟，当此时房间里的人数（进的人数）<出的人数时，才需要计数

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <queue>#include <string>#include <map>#include <stack>#include <list>#include <set>using namespace std;int main(){    int t,n,x,y,cnt,in;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        in = cnt = 0;        for(int i = 0; i < n; i++)        {            scanf("%d%d",&x,&y);            in += x;            if(in < y)            {                cnt += (y-in);                in += (y-in);            }            in -= y;        }        printf("%d\n",cnt);    }    return 0;}