題目鏈接

Problem Description

One day, Dudu, the most clever boy, heard of ACM/ICPC, which is a very interesting game. He wants to take part in the game. But as we all know, you can’t get good result without teammates.So, he needs to select two classmates as his teammates. In this game, the IQ is very important, if you have low IQ you will WanTuo. Dudu’s IQ is a given number k. We use an integer v[i] to represent the IQ of the ith classmate. The sum of new two teammates’ IQ must more than Dudu’s IQ.For some reason, Dudu don’t want the two teammates comes from the same class.Now, give you the status of classes, can you tell Dudu how many ways there are.

Input

There is a number T shows there are T test cases below. (T≤20T≤20)For each test case , the first line contains two integers, n and k, which means the number of class and the IQ of Dudu. n ( 0≤n≤10000≤n≤1000 ), k( 0≤k<2310≤k<231 ).Then, there are n classes below, for each class, the first line contains an integer m, which means the number of the classmates in this class, and for next m lines, each line contains an integer v[i], which means there is a person whose iq is v[i] in this class. m( 0≤m≤1000≤m≤100 ), vi

Output

For each test case, output a single integer.

Sample Input

1 3 1 1 2 1 2 2 1 1

Sample Output

5

AC

• 剛開始寫的時候是開了3個for TLE
• 先將所有IQ存在一起，然后二分找每個數字可能的情況，然后在同一個班級里找可能的情況，消除不符合條件的
• 手寫二分超時。。。

#include <iostream> #include <stdio.h> #include <map> #include <vector> #include <algorithm> #define N 100005 #define ll long long using namespace std; int a[1004][104]; int b[1000005]; int main() { #ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin); #endifint t;scanf("%d", &t);while (t--) {int n, k;scanf("%d%d", &n, &k);int now = 0;for (int i = 0; i < n; ++i) {scanf("%d", &a[i][0]);for (int j = 1; j <= a[i][0]; ++j) {scanf("%d", &a[i][j]);b[now++] = a[i][j];}sort(a[i] + 1, a[i] + 1 + a[i][0]);}sort(b, b + now);ll ans = 0;// 統計所有人的可能情況 for (int i = 0; i < now - 1; ++i) {ans += now - (upper_bound(b + i + 1, b + now, k - b[i]) - b);} // 減去每個班級重復的 for (int i = 0; i < n; ++i) {for (int j = 1; j < a[i][0]; ++j) {ans -= a[i][0] - (upper_bound(a[i] + 1 + j, a[i] + a[i][0] + 1, k - a[i][j]) - a[i]) + 1;}}printf("%lld\n", ans);}return 0; } #include <iostream> #include <stdio.h> #include <map> #include <vector> #include <algorithm> #define N 100005 #define ll long long using namespace std; int a[1004][104]; int b[1000005]; int main() { #ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin); #endifint t;scanf("%d", &t);while (t--) {int n, k;scanf("%d%d", &n, &k);int now = 0;for (int i = 0; i < n; ++i) {scanf("%d", &a[i][0]);for (int j = 1; j <= a[i][0]; ++j) {scanf("%d", &a[i][j]);b[now++] = a[i][j];}sort(a[i] + 1, a[i] + 1 + a[i][0]);}sort(b, b + now);ll ans = 0;for (int i = 0; i < now - 1; ++i) {int l = i + 1, r = now - 1;// 找到第一個大于 while (l <= r) {int mid = (l + r) / 2;if (b[mid] > k - b[i])r = mid - 1;elsel = mid + 1; }ans += now - l;} for (int i = 0; i < n; ++i) {for (int j = 1; j < a[i][0]; ++j) {int l = j + 1, r = a[i][0];// 找到第一個大于 while (l <= r) {int mid = (l + r) / 2;if (a[i][mid] > k - a[i][j])r = mid - 1;elsel = mid + 1;}ans -= a[i][0] - l + 1;}}printf("%lld\n", ans);}return 0; }

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