hdu 5101 Select(樹狀數組)，hdu5101

hdu 5101 Select(樹狀數組)，hdu5101

題目連結：hdu5101 Select

題目大意：N和K，給定若干組數，要從從不同組中選出連個數和大於K，問說有多少種組成方案。

解題思路：樹狀數組維護，將所有的數離散化掉對應成樹狀數組的下標，每次先計算一組，然後再將該組的元素插入到

樹狀數組中。

#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;}