題面
題解
通過一句經典的話“最大值的最小值” 我判斷它是二分題,
不難發現,整個圖形中兩個省的分界線是一條單調不遞減或單調不遞增的折線。
而且,越到後來它的最大值只會越來越大,最小值只會越來越小,極差只會越來越大。
所以如果我們把ans上界定下來了,我們就可以貪心的讓它其中一個區域的值在 [maxa - ans , maxa] 的灰色地帶發展,然後判斷另一個區域是否合法。由於另一個區域的點數越少越好,所以在灰色地帶要儘量發展得廣。
我們只能先定義一個 p 表示當前行兩省的分界, 然後在第一行儘量擴展得遠,下一行在 [0 , p]之間擴展得遠,這樣不僅可以讓另一個區域的點更少,而且可以讓下一行有更大的發展空間。
其它的貪心方法就不行了,所以分別行數從小到大和從大到小各枚舉一次,然後做個左右對稱再來。
CODE
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<set>
#define LL long long
using namespace std;
inline LL read() {
LL f = 1,x = 0;char s = getchar();
while(s < '0' || s > '9') {if(s == '-')f = -1;s = getchar();}
while(s >= '0' && s <= '9'){x = x * 10 + s - '0';s = getchar();}
return x * f;
}
LL n,m,i,j,s,o,k,flag = 0;
LL a[2005][2005];
int b[2005],c[2005];
int x1 = 1,x2 = 1,y1 = 1,y2 = 1;
LL maxl[2005][2005];
LL maxr[2005][2005];
LL minl[2005][2005];
LL minr[2005][2005];
LL ans = 1e18;
bool check(LL mid) {
LL minn = a[x1][y1] - mid;
// cout<<minn<<endl;
LL p = m,mi1 = 0x7f7f7f7f,mi2 = 0x7f7f7f7f,ma1=0,ma2=0;
for(int i = 1;i <= n;i ++) {
int pp = p;
for(int j = 0;j <= pp;j ++) {
if(minl[i][j] >= minn) {
p = j;
}
}
mi1 = min(mi1,minl[i][p]);
ma1 = max(ma1,maxl[i][p]);
mi2 = min(mi2,minr[i][p + 1]);
ma2 = max(ma2,maxr[i][p + 1]);
}
// cout<<"ok1"<<endl;
if(max(ma1 - mi1,ma2 - mi2) <= mid) return 1;
p = m,mi1 = 0x7f7f7f7f,mi2 = 0x7f7f7f7f,ma1=0,ma2=0;
for(int i = n;i > 0;i --) {
int pp = p;
for(int j = 0;j <= pp;j ++) {
if(minl[i][j] >= minn) {
p = j;
}
}
mi1 = min(mi1,minl[i][p]);
ma1 = max(ma1,maxl[i][p]);
mi2 = min(mi2,minr[i][p + 1]);
ma2 = max(ma2,maxr[i][p + 1]);
}
// cout<<"ok2"<<endl;
if(max(ma1 - mi1,ma2 - mi2) <= mid) return 1;
p = 0,mi1 = 0x7f7f7f7f,mi2 = 0x7f7f7f7f,ma1=0,ma2=0;
for(int i = 1;i <= n;i ++) {
int pp = p;
for(int j = m;j >= pp;j --) {
if(minr[i][j + 1] >= minn) {
p = j;
}
}
mi1 = min(mi1,minl[i][p]);
ma1 = max(ma1,maxl[i][p]);
mi2 = min(mi2,minr[i][p + 1]);
ma2 = max(ma2,maxr[i][p + 1]);
}
// cout<<"ok3"<<endl;
if(max(ma1 - mi1,ma2 - mi2) <= mid) return 1;
p = 0,mi1 = 0x7f7f7f7f,mi2 = 0x7f7f7f7f,ma1=0,ma2=0;
for(int i = n;i > 0;i --) {
int pp = p;
for(int j = m;j >= pp;j --) {
if(minr[i][j + 1] >= minn) {
p = j;
}
}
// printf("p---%d\n",p);
mi1 = min(mi1,minl[i][p]);
ma1 = max(ma1,maxl[i][p]);
mi2 = min(mi2,minr[i][p + 1]);
ma2 = max(ma2,maxr[i][p + 1]);
}
// cout<<"ok4"<<endl;
if(max(ma1 - mi1,ma2 - mi2) <= mid) return 1;
return 0;
}
LL solve(LL l,LL r) {
// printf("%lld %lld\n",l,r);
if(l >= r - 1) {
if(check(l)) return l;
return r;
}
LL mid = (l + r) / 2;
if(check(mid)) return solve(l,mid);
return solve(mid,r);
}
int main() {
n = read();m = read();
for(int i = 1;i <= n;i ++) {
minl[i][0] = 1e18;
for(int j = 1;j <= m;j ++) {
a[i][j] = read();
if(a[i][j] > a[x1][y1]) x1 = i,y1 = j;
if(a[i][j] < a[x2][y2]) x2 = i,y2 = j;
minl[i][j] = min(minl[i][j - 1],a[i][j]);
maxl[i][j] = max(maxl[i][j - 1],a[i][j]);
}
minr[i][m + 1] = 1e18;
for(int j = m;j > 0;j --) {
minr[i][j] = min(minr[i][j + 1],a[i][j]);
maxr[i][j] = max(maxr[i][j + 1],a[i][j]);
}
// for(int j = 1;j <= m;j ++) {
// printf("%lld ",minr[i][j]);
// }putchar('\n');
}
printf("%lld\n",solve(0,a[x1][y1] - a[x2][y2]));
return 0;
}