Day3-P - Matrix POJ3685
Given a N × N matrix A, whose element in the i-th row and j-th column Aij is an number that equals i2 + 100000 × i + j2 - 100000 × j + i × j, you are to find the M-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
12 1 1 2 1 2 2 2 3 2 4 3 1 3 2 3 8 3 9 5 1 5 25 5 10
Sample Output
3 -99993 3 12 100007 -199987 -99993 100019 200013 -399969 400031 -99939
思路:观察列递增,行递减,枚举列进行二分算出第M大,然后二分枚举X是第几大,二分套二分,代码如下:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
typedef long long LL; const LL INF = 0x3f3f3f3f3f3f3; LL N, M; LL calculate(LL i, LL j) { return i * i + 100000 * i + j * j - 100000 * j + i * j; } LL check(LL num) { LL sum = 0, l, r, mid; for (int i = 1; i <= N; ++i) { l = 1, r = N; while(l <= r) { mid = (r + l) >> 1; if(calculate(mid,i) > num) { r = mid - 1; } else l = mid + 1; } sum += r; } return sum; } int main() { LL T; scanf("%I64d", &T); while(T--) { scanf("%I64d%I64d", &N, &M); LL l = -INF, r = INF, mid; while(l <= r) { mid = (r + l) >> 1; if(check(mid) < M) { l = mid + 1; } else r = mid - 1; } printf("%I64d\n", l); } return 0; }
小结:如何寻找二分的答案,如果mid是所求答案,那么check(mid)为true,更新r的值为mid-1,之后的check都是false,从而更新l,最后l变为mid,输出mid。
比如>=的情况就是true的情况,所以一般答案都是非>=的那一侧。
补:
在二分时,注意满足的条件是满足题意还是不满足题意,New如下:(一般来说ans都在等于的那一侧,但是哪一侧是l哪一侧是r需要判断
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
typedef long long LL; LL N, M, mid, ans; LL calculate(LL i, LL j) { return i * i + 100000 * i + j * j - 100000 * j + i * j; } LL check(LL x) { LL l, r, mid, sum = 0, ans; for(int i = 1; i <= N; ++i) { l = 1, r = N, ans = 0; while(l <= r) { mid = (r + l) >> 1; if(calculate(mid, i) <= x) { ans = mid; l = mid + 1; } else { r = mid - 1; } } sum += ans; } return sum; } const LL INF = 0x3f3f3f3f3f3f3; int main() { int T; scanf("%d", &T); while(T--) { scanf("%I64d%I64d", &N, &M); //LL l = calculate(1, N), r = calculate(N, 1); LL l = -INF, r = INF; while(l <= r) { mid = (r + l) >> 1; if(check(mid) < M) { l = mid + 1; } else { ans = mid; r = mid - 1; } } printf("%I64d\n", ans); } return 0; }