http://acm.hdu.edu.cn/showproblem.php?pid=5101
题目大意：给n个集合，求从两个不同集合里面各取一个数使得它们的和大于给定数的方案数。
ans=从所有数里面取两个数的方案数-从每个集合里面取两个数的方案数（这是关键）
如果不转换也可以这么做，离散一下，然后树状数组统计也行，具体见代码。
代码一：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <map> 7 #include <vector> 8 #include <stack> 9 #include <string> 10 #include <ctime> 11 #include <queue> 12 #define mem0(a) memset(a, 0, sizeof(a)) 13 #define mem(a, b) memset(a, b, sizeof(a)) 14 #define lson l, m, rt << 1 15 #define rson m + 1, r, rt << 1 | 1 16 #define eps 0.0000001 17 #define lowbit(x) ((x) & -(x)) 18 #define memc(a, b) memcpy(a, b, sizeof(b)) 19 #define x_x(a) ((a) * (a)) 20 #define LL long long 21 #define DB double 22 #define pi 3.14159265359 23 #define MD 10000007 24 #define INF (int)1e9 25 #define max(a, b) ((a) > (b)? (a) : (b)) 26 using namespace std; 27 int b[110000]; 28 LL solve(int a[], int n, int x) 29 { 30 LL ans = 0; 31 for(int i = 1; i <= n; i++) { 32 int tmp = upper_bound(a + 1, a + 1 + n, x - a[i]) - a; 33 ans += n - tmp + 1; 34 } 35 return ans; 36 } 37 int main() 38 { 39 //freopen("input.txt", "r", stdin); 40 //freopen("output.txt", "w", stdout); 41 int n, k, T; 42 cin>> T; 43 while(T--) { 44 cin>> n>> k; 45 LL ans1 = 0, ans2 = 0; 46 int tot = 0; 47 for(int i = 1; i <= n; i++) { 48 int m, a[110]; 49 cin>> m; 50 for(int j = 1; j <= m; j++) { 51 scanf("%d", &a[j]); 52 b[++tot] = a[j]; 53 } 54 sort(a + 1, a + 1 + m); 55 ans1 += solve(a, m, k); 56 } 57 sort(b + 1, b + 1 + tot); 58 ans2 = solve(b, tot, k); 59 cout<< (ans2 - ans1) / 2<< endl; 60 } 61 return 0; 62 }
代码二（树状数组）：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <map> 7 #include <vector> 8 #include <stack> 9 #include <string> 10 #include <ctime> 11 #include <queue> 12 #define mem0(a) memset(a, 0, sizeof(a)) 13 #define mem(a, b) memset(a, b, sizeof(a)) 14 #define lson l, m, rt << 1 15 #define rson m + 1, r, rt << 1 | 1 16 #define eps 0.0000001 17 #define lowbit(x) ((x) & -(x)) 18 #define memc(a, b) memcpy(a, b, sizeof(b)) 19 #define x_x(a) ((a) * (a)) 20 #define LL long long 21 #define DB double 22 #define pi 3.14159265359 23 #define MD 10000007 24 #define INF (int)1e9 25 #define max(a, b) ((a) > (b)? (a) : (b)) 26 using namespace std; 27 map<int, int> hash; 28 int nn, tot, arr[220000], arr0[220000], c[220000], a[1200][120], m[1200]; 29 void init() 30 { 31 sort(arr + 1, arr + 1 + tot); 32 arr0[1] = arr[1]; 33 nn = 1; 34 hash[arr0[1]] = 1; 35 for(int i = 2; i <= tot; i++) { 36 if(arr[i] != arr[i - 1]) { 37 arr0[++nn] = arr[i]; 38 hash[arr[i]] = nn; 39 } 40 } 41 } 42 void update(int p, int x) 43 { 44 while(p <= nn) { 45 c[p] += x; 46 p += lowbit(p); 47 } 48 } 49 void insert(int a[], int n) 50 { 51 for(int i = 1; i <= n; i++) { 52 int tmp = hash[a[i]]; 53 update(tmp, 1); 54 } 55 } 56 int sum(int p) 57 { 58 int ans = 0; 59 while(p) { 60 ans += c[p]; 61 p -= lowbit(p); 62 } 63 return ans; 64 } 65 int query(int L, int R) 66 { 67 return sum(R) - sum(L - 1); 68 } 69 int find(int x) 70 { 71 int l = 1, r = nn; 72 while(l < r) { 73 int m = (l + r) >> 1; 74 if(arr0[m] > x) r = m; 75 else l = m + 1; 76 } 77 return l; 78 } 79 int main() 80 { 81 //freopen("input.txt", "r", stdin); 82 //freopen("output.txt", "w", stdout); 83 int T; 84 cin>> T; 85 while(T--) { 86 hash.clear(); 87 mem0(c); 88 int n, k; 89 cin>> n>> k; 90 tot = 0; 91 for(int i = 1; i <= n; i++) { 92 cin>> m[i]; 93 for(int j = 1; j <= m[i]; j++) { 94 scanf("%d", &a[i][j]); 95 arr[++tot] = a[i][j]; 96 } 97 } 98 init(); 99 insert(a[1], m[1]); 100 LL ans = 0; 101 for(int i = 2; i <= n; i++) { 102 for(int j = 1; j <= m[i]; j++) { 103 int tmp = find(k - a[i][j]); 104 ans += query(tmp, nn); 105 } 106 insert(a[i], m[i]); 107 } 108 cout<< ans<< endl; 109 } 110 return 0; 111 }