[HAOI2007]理想的正方形 BZOJ1047 二維RMQ
阿新 • • 發佈：2019-02-10
tdi deque 一行 brush tor 行為 har pre 空格
題目描述
有一個a*b的整數組成的矩陣，現請你從中找出一個n*n的正方形區域，使得該區域所有數中的最大值和最小值的差最小。
輸入輸出格式
輸入格式：第一行為3個整數，分別表示a,b,n的值
第二行至第a+1行每行為b個非負整數，表示矩陣中相應位置上的數。每行相鄰兩數之間用一空格分隔。
輸出格式：僅一個整數，為a*b矩陣中所有“n*n正方形區域中的最大整數和最小整數的差值”的最小值。
輸入輸出樣例
輸入樣例#1： 復制5 4 2 1 2 5 6 0 17 16 0 16 17 2 1 2 10 2 1 1 2 2 2輸出樣例#1： 復制
1
說明
問題規模
（1）矩陣中的所有數都不超過1,000,000,000
（2）20%的數據2<=a,b<=100,n<=a,n<=b,n<=10
（3）100%的數據2<=a,b<=1000,n<=a,n<=b,n<=100
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 700005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == ‘-‘) f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
int a, b, n;
int Log;
int maxx[1103][1103];
int minn[1103][1103];
int mx[1101][1101];
int query(int x, int y) {
int Max = -inf, Min = inf;
Max = max(maxx[x][y], max(maxx[x + n - (1 << Log)][y + n - (1 << Log)], max(maxx[x + n - (1 << Log)][y], maxx[x][y + n - (1 << Log)])));
Min = min(minn[x][y], min(minn[x + n - (1 << Log)][y + n - (1 << Log)], min(minn[x + n - (1 << Log)][y], minn[x][y + n - (1 << Log)])));
return Max - Min;
}
int main()
{
// ios::sync_with_stdio(0);
a = rd(); b = rd(); n = rd();
for (int i = 1; i <= a; i++) {
for (int j = 1; j <= b; j++) {
mx[i][j] = rd();
maxx[i][j] = minn[i][j] = mx[i][j];
}
}
for (Log = 0; (1 << (Log + 1) <= n); Log++);
for (int k = 0; k < Log; k++) {
for (int i = 1; i + (1 << k) <= a; i++) {
for (int j = 1; j + (1 << k) <= b; j++) {
maxx[i][j] = max(maxx[i][j], max(maxx[i + (1 << (k))][j + (1 << (k))], max(maxx[i][j + (1 << k)], maxx[i + (1 << k)][j])));
minn[i][j] = min(minn[i][j], min(minn[i + (1 << k)][j + (1 << k)], min(minn[i + (1 << k)][j], minn[i][j + (1 << k)])));
}
}
}
ll ans = 9999999999;
for (int i = 1; i <= a - n + 1; i++) {
for (int j = 1; j <= b - n + 1; j++) {
ans = min(ans, 1ll * query(i, j));
}
}
printf("%d\n", ans);
return 0;
}
[HAOI2007]理想的正方形 BZOJ1047 二維RMQ