题目链接：hdu5101 Select

题目大意：N和K，给定若干组数，要从从不同组中选出连个数和大于K，问说有多少种组成方案。

解题思路：树状数组维护，将所有的数离散化掉对应成树状数组的下标，每次先计算一组，然后再将该组的元素插入到

树状数组中。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 1005;
const int maxm = 105;
const int maxs = 1e5 + 1;
#define lowbit(x) ((x)&(-x))

int N, M, cn[maxn], fenw[maxs + 5];
ll K, clas[maxn][maxm];
ll idx[maxs + 5], tmp[maxs + 5];

inline void add(int x, int d) {
    while (x <= maxs) {
        fenw[x] += d;
        x += lowbit(x);
    }
}

inline int sum(int x) {
    int ret = 0;
    while (x) {
        ret += fenw[x];
        x -= lowbit(x);
    }
    return ret;
}

void init () {
    int n = M = 0;
    scanf("%d%I64d", &N, &K);
    memset(cn, 0, sizeof(cn));
    memset(fenw, 0, sizeof(fenw));

    for (int i = 1; i <= N; i++) {
        scanf("%d", &cn[i]);
        for (int j = 0; j < cn[i]; j++) {
            scanf("%I64d", &clas[i][j]);
            tmp[n++] = clas[i][j];
        }
    }

    sort(tmp, tmp + n);
    idx[M++] = tmp[0];
    for (int i = 1; i < n; i++) {
        if (tmp[i] != tmp[i-1])
            idx[M++] = tmp[i];
    }
}

inline int find (ll x) {
    return (lower_bound(idx, idx + M, x) - idx) + 1;
}

void insert(int d) {
    for (int i = 0; i < cn[d]; i++) {
        int pos = find(clas[d][i]);
        add(pos, 1);
    }
}

ll query(int d) {
    ll ret = 0;
    for (int i = 0; i < cn[d]; i++) {
        int pos = find(K - clas[d][i] + 1);
        ret += sum(maxs) - sum(pos-1);
    }
    return ret;
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        init();

        ll ans = 0;
        insert(1);
        for (int i = 2; i <= N; i++) {
            ans += query(i);
            insert(i);
        }
        printf("%I64d\n", ans);
    }
    return 0;
}

hdu 5101 Select(树状数组)