Poj 3685（经典二分）

Matrix

Time Limit: 6000MS Memory Limit: 65536KTotal Submissions: 4150 Accepted: 1007

Description

Given a N × N matrix A, whose element in the i-th row andj-th columnA_ij is an number that equals i² + 100000 ×i +j² - 100000 × j + i × j, you are to find theM-th smallest element in the matrix.

Input

The first line of input is the number of test case.
For each test case there is only one line contains two integers, N(1 ≤ N ≤ 50,000) and M(1 ≤ M ≤ N × N). There is a blank line before each test case.

Output

For each test case output the answer on a single line.

Sample Input

121 12 12 22 32 43 13 23 83 95 15 255 10

Sample Output

3-99993312100007-199987-99993100019200013-399969400031-99939

Source

POJ Founder Monthly Contest – 2008.08.31, windy7926778

题目意思是给出一个矩阵，这个矩阵中各个元素的值由给定的公式确定。即：i² + 100000 ×i +j² - 100000 × j + i × j。开始打了个10*10的表，以为矩阵元素得大小顺序是有规律的，交了后wa了，发现还是不能乱搞。从该式子可以看出，结果随着i单调递增，但对j呢？求导后可发现不是单调递增的，所以找规律就是错的。求第k大元素嘛，还是考虑二分。二分的关键是怎么去判断。因为j无单调性可言，那么就枚举j，二分找出每列的小于mid的个数cntx，和小于等于mid的个数cnty。总复杂度就是logINF*n*logn，是可以接受的。然后根据相应逻辑判断即可。这道题，就是二分里再套二分，还是蛮有意思的。

#include<cstdio>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<iostream>#include<map>using namespace std;typedef long long LL;const int maxn = 1000 + 5;const LL INF = 1LL<<40;LL n,m;LL cal(LL i,LL j){    return  i*i + 100000 * i + j*j - 100000 * j + i * j;}int can(LL x){    LL cntx = 0,cnty = 0;    for(LL i = 1;i <= n;i++){        LL l = 1,r = n;        LL tem = -1;        while(l <= r){            LL mid = l+(r-l)/2;            if(cal(mid,i) < x){                l = mid+1;                tem = mid;            }            else {                r = mid-1;            }        }        if(tem != -1){            cntx += tem;        }    }    for(LL i = 1;i <= n;i++){        LL l = 1,r = n;        LL tem = -1;        while(l <= r){            LL mid = l+(r-l)/2;            if(cal(mid,i) <= x){                l = mid+1;                tem = mid;            }            else {                r = mid-1;            }        }        if(tem != -1){            cnty += tem;        }    }    if(cnty < m){        return 1;    }    else if(cntx > m-1){        return -1;    }    else{        return 0;    }}int main(){    int t;    scanf("%d",&t);    while(t--){        scanf("%I64d%I64d",&n,&m);        LL l = -INF,r = INF;        LL ans;        while(l <= r){            LL mid = l+(r-l)/2;            int tag = can(mid);            if(tag == 1){                l = mid+1;            }            else if(tag == -1){                r = mid-1;            }            else{                ans = mid;                break;            }        }        printf("%I64d\n",ans);    }    return 0;}