牛客2017校招真题在线编程 分田地（二维前缀和+二分

题目链接

思路：思路很惊奇啊，首先二分答案，得到mid，再看分到的部分是否16个部分都大于等于mid，如果有小于mid的，则mid还要取更小一点。再从一维出发，三层for切三刀，再从另外一维出发，切一刀就好了，之前我还在想切一刀如何鉴定16个部分，因为如果每次从（0，0）作为最左上角只能鉴定4个部分，而且还是最左边的4个部分，后来看了他们的写法，把上一个点的终点作为下一个点的左上点的起点既而16个部分都能鉴定了。另外则是二维前缀和的处理。

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.
///             __.'              ~.   .~              `.__
///           .'//                  \./                  \\`.
///        .'//                     |                     \\`.
///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.
///     .'//.-"                 `-.  |  .-'                 "-.\\`.
///   .'//______.============-..   \ | /   ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma GCC optimize("Ofast")
#pragma comment(linker, "/STACK:102400000,102400000")
#pragma GCC target(sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx) 
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair <int, int> ii;
const int INF=-1u>>1; 
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e6+10;
const int maxx=1e3+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

int n,m; 
int a[80][80];
char s[80][80];
int sum[80][80];

int calc(int x1,int y1,int x2,int y2){
    return sum[x2][y2]-sum[x2][y1]-sum[x1][y2]+sum[x1][y1];
}
int check(int mid){
	FOR(1,n-3,i){
		FOR(i+1,n-2,ii){
			FOR(ii+1,n-1,iii){
				int xx=0,cnt=0;
				//cout<<i<<" "<<ii<<" "<<iii<<endl;
				FOR(1,m,j){
					if(calc(0,xx,i,j)>=mid&&
					calc(i,xx,ii,j)>=mid&&
					calc(ii,xx,iii,j)>=mid&&
					calc(iii,xx,n,j)>=mid){
						xx=j;
						cnt++;
					}
				}
				if(cnt>=4){
					return 1;
				}
			}
		}
	}
	return 0;
}
void solve(){
	W(cin>>n>>m){
		FOR(1,n,i)
			cin>>s[i]+1;
		FOR(1,n,i){
			FOR(1,m,j){
				a[i][j]=s[i][j]-'0';
			}
		}
		me(sum,0);
		FOR(1,n,i){
			FOR(1,m,j){
				sum[i][j]=sum[i][j-1]+sum[i-1][j]-sum[i-1][j-1]+a[i][j];
			}
		}
		int l=0,r=sum[n][m],ans=0;
		W(l<=r){
			int mid=(l+r)/2;
			if(check(mid)){
				l=mid+1;
				ans=mid;
			}
			else{
				r=mid-1;
			}
		}
		print(ans);
	}
}
int main(){
    //freopen( "in.txt" , "r" , stdin );
    //freopen( "data.txt" , "w" , stdout );
    int t=1;
    //init();
    //s_1(t);
    for(int cas=1;cas<=t;cas++){
        //printf("Case #%d: ",cas);
        solve();
    }
}