zoukankan html css js c++ java

UPC 2224 Boring Counting ★(山东省第四届ACM程序设计竞赛 tag:线段树)

[题意]给定一个长度为N的数列，M个询问区间[L,R]内大于等于A小于等于B的数的个数. [题目链接]http://acm.upc.edu.cn/problem.php?id=2224 省赛的时候脑抽想了10min没想出来就看别的题去了= =，赛后又想了10min想出来了并且1Y。。。真嫌弃自己= =。。。 [分析]如果做过询问区间[L,R]内小于H的个数的那道线段树题(HDU 4417 2012年杭州赛区网络赛)那么这题就好想了。那在这里再说一下做法：｛将所有的询问离线读入之后，按H从小到大排序。对于所有的区间数也按从小到大排序，然后根据查询的H，将比H小的点加入到线段树，然后就是一个区间和｝。那么这题也就简单了。。。按上面方法分别求出区间[L,R]内大于等于A的个数num1和小于等于B的个数num2，然后res = num1 + num2 - [L,R].

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define MID(x,y) ((x+y)>>1)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
 
const int MAXN = 50002;
struct Segment_Tree{
    int sum[MAXN<<2];
    void build(){
        mem(sum, 0);
    }
    void pushup(int rt){
        sum[rt] = sum[rt<<1] + sum[rt<<1|1];
    }
    void update(int p, int v, int l, int r, int rt){
        if (l == r){
            sum[rt] = v;
            return ;
        }
        int mid = MID(l, r);
        if (p <= mid) update(p, v, l, mid, rt<<1);
        else    update(p, v, mid+1, r, rt<<1|1);
        pushup(rt);
    };
    int query(int l, int r, int L, int R, int rt){
        if (l <= L && R <= r){
            return sum[rt];
        }
        int mid = MID(L,R);
        int ans = 0;
        if (l <= mid)   ans += query(l, r, L, mid, rt<<1);
        if (mid < r)    ans += query(l, r, mid+1, R, rt<<1|1);
        return ans;
    }
}S;
struct num{
    int value;
    int position;
}A[MAXN];
struct queries{
    int l, r;
    int a, b;
    int num_greater_a;
    int num_less_b;
    int res;
    int id;
}q[MAXN];
bool cmp(num n1, num n2){
    return n1.value < n2.value;
}
bool cmp1(queries n1, queries n2){
    return n1.a > n2.a;
}
bool cmp2(queries n1, queries n2){
    return n1.b < n2.b;
}
bool cmpid(queries n1, queries n2){
    return n1.id < n2.id;
}
int main(){
    int t;
    scanf("%d", &t);
    for (int o = 1; o <= t; o ++){
        int n, m;
        scanf("%d %d", &n, &m);
        for (int i = 0; i < n; i ++){
            scanf("%d", &A[i].value);
            A[i].position = i + 1;
        }
        for (int i = 0; i < m; i ++){
            scanf("%d %d %d %d", &q[i].l, &q[i].r, &q[i].a, &q[i].b);
            q[i].id = i;
        }
        sort(A, A+n, cmp);
        sort(q, q+m, cmp1);
        int j = n - 1;
        S.build();
        for (int i = 0; i < m; i ++){
            while(j >= 0 && A[j].value >= q[i].a){
                S.update(A[j].position, 1, 1, n, 1);
                j --;
            }
            q[i].num_greater_a = S.query(q[i].l, q[i].r, 1, n, 1);
        }
        sort(q, q+m, cmp2);
        j = 0;
        S.build();
        for (int i = 0; i < m; i ++){
            while(j < n && A[j].value <= q[i].b){
                S.update(A[j].position, 1, 1, n, 1);
                j ++;
            }
            q[i].num_less_b = S.query(q[i].l, q[i].r, 1, n, 1);
            q[i].res = q[i].num_greater_a + q[i].num_less_b - (q[i].r - q[i].l + 1);
        }
        sort(q, q+m, cmpid);
        printf("Case #%d:
", o);
        for (int i = 0; i < m; i ++){
            printf("%d
", q[i].res);
        }
    }
    return 0;
}
 
/**************************************************************
    Problem: 2224
    User: AbandonZHANG
    Language: C++
    Result: Accepted
    Time:1052 ms
    Memory:4016 kb
****************************************************************/

查看全文

相关阅读:
hdu 1108 最小公倍数
 hdu 1106 排序
 hdu 1097 A hard puzzle
hdu 1076 An Easy Task
hdu 1064 Financial Management
hdu 1061 Rightmost Digit
hdu 1050 Moving Tables
hdu 1060 Leftmost Digit
hdu 1049 Climbing Worm
hdu1104

原文地址：https://www.cnblogs.com/AbandonZHANG/p/4114251.html