HDU 4417 Super Mario （树状数组+离线处理）(划分树+二分答案)

题意： 给定1--n区间，有q个询问，询问l,r,k表示区间[l,r]小于等于k的数的个数

思路： 可以用划分树（求区间第k大值）变形一下，来求小于等于k的个数，但是此题直接离线处理询问高效的多。

首先将1--n区间的值记录位置，从小到大排序，每个询问按照k值从小到大排序，然后从小到大开始，根据查询的H，将满足条件的的点插入，计数+1，然后就是求区间和。

#include <iostream>#include <algorithm>#include <cmath>#include<functional>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <vector>#include <set>#include <queue>#include <stack>#include <climits>//形如INT_MAX一类的#define MAX 100001using namespace std;inline void RD(int &ret) {    char c;    do {        c = getchar();    } while(c < '0' || c > '9') ;    ret = c - '0';    while((c=getchar()) >= '0' && c <= '9')        ret = ret * 10 + ( c - '0' );}void OT(int a) {    if(a >= 10)OT(a / 10);    putchar(a % 10 + '0');}int n,q;struct node {    int v,id;} a[MAX];int c[MAX];struct QES {    int l,r,h,id;} qes[MAX];int ans[MAX];bool cmp(const node &a, const node &b) {    return a.v < b.v;}bool cmp2(const QES &a, const QES &b) {    return a.h < b.h;}int lowbit(int x) {    return x & (-x);}void update(int x,int va) {    while(x <= n) {        c[x] += va;        x += lowbit(x);    }}int query(int x) {    int ans = 0;    while(x > 0) {        ans += c[x];        x -= lowbit(x);    }    return ans;}int main() {    int T;    cin >> T;    int ca = 1;    while(T--) {        RD(n); RD(q);        memset(c,0,sizeof(c));        for(int i=1; i<=n; i++) {            RD(a[i].v);            a[i].id = i;        }        sort(a+1,a+1+n,cmp);        for(int i=0; i<q; i++) RD(qes[i].l),RD(qes[i].r),RD(qes[i].h),qes[i].id = i;        sort(qes,qes+q,cmp2);        int order = 1;        for(int i=0; i<q; i++) {            while(a[order].v <= qes[i].h && order <= n) {                update(a[order].id,1);                order ++;            }            ans[qes[i].id] = query(qes[i].r+1) - query(qes[i].l);        }        printf("Case %d:\n",ca++);        for(int i=0; i<q; i++)OT(ans[i]),puts("");    }    return 0;}

划分树：

要求的是小于等于k的个数，可以二分该个数mid，找到的也是第mid小的值，与k大小比较，直到得到答案

#include <iostream>#include <algorithm>#include <cmath>#include<functional>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <vector>#include <set>#include <queue>#include <stack>#include <climits>//形如INT_MAX一类的#define MAX 100005#define INF 0x7FFFFFFF#define L(x) x << 1#define R(x) x << 1 | 1using namespace std;struct Seg_Tree {    int l,r,mid;} tr[MAX*4];int sorted[MAX];int lef[20][MAX];int val[20][MAX];void build(int l,int r,int step,int x) {    tr[x].l = l;    tr[x].r = r;    tr[x].mid = (l + r) >> 1;    if(tr[x].l == tr[x].r) return ;    int mid = tr[x].mid;    int lsame = mid - l + 1;//lsame表示和val_mid相等且分到左边的    for(int i = l ; i <= r ; i ++) {        if(val[step][i] < sorted[mid]) {            lsame --;//先假设左边的数(mid - l + 1)个都等于val_mid,然后把实际上小于val_mid的减去        }    }    int lpos = l;    int rpos = mid + 1;    int same = 0;    for(int i = l ; i <= r ; i ++) {        if(i == l) {            lef[step][i] = 0;//lef[i]表示[ tr[x].l , i ]区域里有多少个数分到左边        } else {            lef[step][i] = lef[step][i-1];        }        if(val[step][i] < sorted[mid]) {            lef[step][i] ++;            val[step + 1][lpos++] = val[step][i];        } else if(val[step][i] > sorted[mid]) {            val[step+1][rpos++] = val[step][i];        } else {            if(same < lsame) {//有lsame的数是分到左边的                same ++;                lef[step][i] ++;                val[step+1][lpos++] = val[step][i];            } else {                val[step+1][rpos++] = val[step][i];            }        }    }    build(l,mid,step+1,L(x));    build(mid+1,r,step+1,R(x));}int query(int l,int r,int k,int step,int x) {    if(l == r) {        return val[step][l];    }    int s;//s表示[l , r]有多少个分到左边    int ss;//ss表示 [tr[x].l , l-1 ]有多少个分到左边    if(l == tr[x].l) {        s = lef[step][r];        ss = 0;    } else {        s = lef[step][r] - lef[step][l-1];        ss = lef[step][l-1];    }    if(s >= k) {//有多于k个分到左边,显然去左儿子区间找第k个        int newl = tr[x].l + ss;        int newr = tr[x].l + ss + s - 1;//计算出新的映射区间        return query(newl,newr,k,step+1,L(x));    } else {        int mid = tr[x].mid;        int bb = l - tr[x].l - ss;//bb表示 [tr[x].l , l-1 ]有多少个分到右边        int b = r - l + 1 - s;//b表示 [l , r]有多少个分到右边        int newl = mid + bb + 1;        int newr = mid + bb + b;        return query(newl,newr,k-s,step+1,R(x));    }}int n,m;int main() {    int T;    cin >> T;    int l,r,k;    int ca = 1;    while(T--) {        scanf("%d%d",&n,&m);        for(int i=1; i<=n; i++) {            scanf("%d",&val[0][i]);            sorted[i] = val[0][i];        }        sort(sorted+1,sorted+1+n);        build(1,n,0,1);        printf("Case %d:\n",ca++);        for(int i=0; i<m; i++) {            scanf("%d%d%d",&l,&r,&k);            l ++;            r ++;            int ll = 1;            int rr = r - l + 1;            int mid,maxx = 0;            while(ll <= rr) {                mid = (ll + rr) >> 1;                if(query(l,r,mid,0,1) <= k) {                    maxx = max(maxx,mid);                    ll = mid + 1;                } else rr = mid - 1;            }            printf("%d\n",maxx);        }    }    return 0;}