Codeforces Round #401 (Div. 2) C. Alyona and Spreadsheet

题目链接：C. Alyona and Spreadsheet

题意：

给你一个n*m的矩阵，现在有k个询问，每次给你一个l，r，问你在[l,r]这行中能否有一列数十非递减的顺序

题解：

用vector来保存矩阵。

对于每一行n*m dp一下最远能达到的范围，然后询问的时候就判断l,r是否在这个范围内。

#include<bits/stdc++.h>
#define pb push_back
#define rs m+1,r,rt<<1|1
#define mst(a,b) memset(a,b,sizeof(a))
#define F(i,a,b) for(int i=a;i<=b;++i)
#define ___ freopen("d:\\acm\\input.txt","r",stdin);
using namespace std;
typedef long long ll;

vector<vector<int> >a;
vector<vector<int> >dp;
vector<int>mx;
int n,m,x;

int main(){
    scanf("%d%d",&n,&m);
    a.resize(n+2),dp.resize(n+2),mx.resize(n+2);
    F(i,0,n)a[i].resize(m+2),dp[i].resize(m+2);
    F(i,1,n)F(j,1,m)scanf("%d",&x),a[i][j]=x;
    F(i,1,m)dp[1][i]=1;
    F(i,1,n)mx[i]=0;
    F(i,2,n)F(j,1,m)
    if(a[i][j]>=a[i-1][j])dp[i][j]=dp[i-1][j]+1;
    else dp[i][j]=1;
    F(i,1,n)F(j,1,m)mx[i]=max(mx[i],dp[i][j]);
    int q;
    scanf("%d",&q);
    while(q--)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        if(y-x+1<=mx[y])puts("Yes");else puts("No");
    }
    return 0;
}