重庆选科网站,香河县做网站,上海阿里巴巴做网站,注册一个电商平台需要多少钱题意#xff1a;给你一个n*m的数阵 对于一行到另一行#xff0c;若存在一列从上到下递减#xff0c;则称之符合题意 The first line of the input contains two positive integers n and m (1 ≤ nm ≤ 100 000) — the number of rows and the number of columns in t…题意给你一个n*m的数阵 对于一行到另一行若存在一列从上到下递减则称之符合题意 The first line of the input contains two positive integers n and m (1 ≤ n·m ≤ 100 000) — the number of rows and the number of columns in the table respectively. Note that your are given a constraint that bound the product of these two integers, i.e. the number of elements in the table. Each of the following n lines contains m integers. The j-th integers in the i of these lines stands for ai, j (1 ≤ ai, j ≤ 109).The next line of the input contains an integer k (1 ≤ k ≤ 100 000) — the number of task that teacher gave to Alyona.The i-th of the next k lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n).
以上是数据范围 a【】来存储每一行的数 b【】来存储每一列能到达的最上行 在用c【】来存储每一行能到达的最上行 #includecstdio
using namespace std;
int a[100005],b[100005],c[100005];
int main()
{int n,m,x,i,j,r,l,k;scanf(%d%d,n,m);for(i1;im;i)b[i]1;for(i1;in;i){c[i]i; for(j1;jm;j){scanf(%d,x);if(xa[j])b[j]i;a[j]x; if(b[j]c[i]) c[i]b[j];}}scanf(%d,k);while(k--){scanf(%d%d,r,l);if(c[l]r)printf(Yes\n);elseprintf(No\n);}return 0;
} 。 思路由于有多次询问所以先打表记录每一行所能到达的最上行即可 转载于:https://www.cnblogs.com/pot-a-to/p/10937035.html
