[HDOJ4417]Super Mario（归并树）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4417

题意：给定n个数和q次询问，每次询问[l,r]区间内不比k大的数字的个数。

直接上归并树，每次查询直接查区间内小于等于k的数的个数，二分即可。

#include <bits/stdc++.h>
using namespace std;

#define lrt rt << 1
#define rrt rt << 1 | 1
const int maxn = 100100;
int n, q, a[maxn];
vector<int> seg[maxn<<2];

void build(int l, int r, int rt) {
    if(l == r) {
        seg[rt].clear();
        seg[rt].push_back(a[l]);
        return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, lrt);
    build(mid+1, r, rrt);
    seg[rt].resize(r-l+1);
    merge(seg[lrt].begin(), seg[lrt].end(), seg[rrt].begin(), seg[rrt].end(), seg[rt].begin());
}

int query(int L, int R, int v, int l, int r, int rt) {
    if(R < l || r < L) return 0;
    if(L <= l && r <= R) return upper_bound(seg[rt].begin(), seg[rt].end(), v) - seg[rt].begin();
    int mid = (l + r) >> 1;
    return query(L, R, v, l, mid, lrt) + query(L, R, v, mid+1, r, rrt);
}

inline bool scan_d(int &num) {
    char in;bool IsN=false;
    in=getchar();
    if(in==EOF) return false;
    while(in!='-'&&(in<'0'||in>'9')) in=getchar();
    if(in=='-'){ IsN=true;num=0;}
    else num=in-'0';
    while(in=getchar(),in>='0'&&in<='9'){
        num*=10,num+=in-'0';
    }
    if(IsN) num=-num;
    return true;
}

inline void out(int x) { 
    if (x > 9) out(x / 10); 
    putchar(x % 10 + '0');
}

int main() {
    // freopen("in", "r", stdin);
    // freopen("out", "w", stdout);
    int T, l, r, k, _ = 1;
    scan_d(T);
    while(T--) {
        scan_d(n); scan_d(q);
        for(int i = 1; i <= n; i++) scan_d(a[i]);
        build(1, n, 1);
        sort(a+1, a+n+1);
        printf("Case %d:
", _++);
        while(q--) {
            scan_d(l); scan_d(r); scan_d(k);
            l++; r++;
            int lo = 1, hi = n;
            int ret = query(l, r, k, 1, n, 1);
            printf("%d
", ret);
        }
    }
    return 0;
}

用数组写了个，快了不少。

#include <bits/stdc++.h>
using namespace std;

#define lrt rt << 1
#define rrt rt << 1 | 1
const int maxn = 100100;
int n, q, a[maxn];
int ll[maxn], rr[maxn];
int seg[20][maxn];

void build(int l, int r, int rt) {
    if(l == r) {
        seg[rt][l] = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, rt+1);
    build(mid+1, r, rt+1);
    // merge(seg[rt+1]+l, seg[rt+1]+mid+1, seg[rt+1]+mid+1, seg[rt+1]+r+1, seg[rt]+l);
    int n1 = mid - l + 1, n2 = r - (mid + 1) + 1;
    int i = 0, j = 0;
    for(int ii = 0; ii < n1; ii++) ll[ii] = seg[rt+1][l+ii];
    for(int ii = 0; ii < n2; ii++) rr[ii] = seg[rt+1][(mid+1)+ii];
    while(i < n1 && j < n2) {
        if(ll[i] <= rr[j]) seg[rt][l++] = ll[i++];
        else seg[rt][l++] = rr[j++];
    }
    while(i < n1) seg[rt][l++] = ll[i++];
    while(j < n2) seg[rt][l++] = rr[j++];
}

int query(int L, int R, int v, int l, int r, int rt) {
    if(r < L || R < l) return 0;
    if(L <= l && r <= R) return upper_bound(seg[rt]+l, seg[rt]+r+1, v) - (seg[rt]+l);
    int mid = (l + r) >> 1;
    return query(L, R, v, l, mid, rt+1) + query(L, R, v, mid+1, r, rt+1);
}

int main() {
    // freopen("in", "r", stdin);
    // freopen("out", "w", stdout);
    int T, l, r, k, _ = 1;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d",&n,&q);
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        build(1, n, 1);
        printf("Case %d:
", _++);
        while(q--) {
            scanf("%d%d%d",&l,&r,&k);
            l++; r++;
            int lo = 1, hi = n;
            int ret = query(l, r, k, 1, n, 1);
            printf("%d
", ret);
        }
    }
    return 0;
}