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

看代码 
一种新的思想 dp
以每行

#include<bits/stdc++.h>using namespace std;const int MAXN=100010;int n,m,a[MAXN],last[MAXN],h[MAXN],k;int main(){scanf("%d%d",&n,&m);for(int i=1;i<=m;++i){scanf("%d",&last[i]);a[i]=1;}h[1]=1;for(int i=2;i<=n;++i){h[i]=MAXN;for(int j=1;j<=m;++j){int t;scanf("%d",&t);if(t<last[j])a[j]=i;h[i]=min(h[i],a[j]);last[j]=t;}}scanf("%d",&k);for(int i=1;i<=k;++i){int l,r;scanf("%d%d",&l,&r);if(h[r]<=l)printf("Yes\n");//h[r]是以r为最右端,使式子成立的最小l; else printf("No\n");}return 0;}