Poj 3685（經典二分）

Matrix

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 4150		Accepted: 1007

Description

Given a N × N matrix A, whose element in the i-th row andj-th column
Aij 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

題目意思是給出一個矩陣，這個矩陣中各個元素的值由給定的公式確定。即：i2 + 100000 ×i +
j2 - 100000 × j + i × j。開始打了個10*10的表，以為矩陣元素得大小順序是有規律的，交了後wa了，發現還是不能亂搞。從該式子可以看出，結果隨著i單調遞增，但對j呢？求導後可發現不是單調遞增的，所以找規律就是錯的。求第k大元素嘛，還是考慮二分。二分的關鍵是怎麼去判斷。因為j無單調性可言，那麼就枚舉j，二分找出每列的小於mid的個數cntx，和小於等於mid的個數cnty。總複雜度就是logINF*n*logn，是可以接受的。然後根據相應邏輯判斷即可。這道題，就是二分裡再套二分，還是蠻有意思的。

#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;}