HDU---4417Super Mario 树状数组 离线操作

题意：给定 n个数，查询 位置L R内 小于x的数有多少个。

对于某一次查询 把所有比x小的数 ”的位置“ 都加入到树状数组中，然后sum(R)-sum(L-1)就是答案，q次查询就要离线操作了，按高度排序。

#include <set>

#include <map>

#include <cmath>

#include <ctime>

#include <queue>

#include <stack>

#include <cctype>

#include <cstdio>

#include <string>

#include <vector>

#include <cstdlib>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

typedef unsigned long long ull;

typedef long long ll;

const int inf = 0x3f3f3f3f;

const double eps = 1e-8;

template <class T>

inline bool scan_d(T &ret)

{

    char c;

    int sgn;

    if(c=getchar(),c==EOF)

    return 0;

    while(c!='-'&&(c<'0'||c>'9'))     c=getchar();

    sgn = (c=='-')?-1:1;

    ret =(c=='-')?0:(c-'0');

    while(c=getchar(),c>='0'&&c<='9')     ret=ret*10+(c-'0');

    ret*=sgn;

    return 1;

}

const int maxn = 1e5+100;

int n,q,c[maxn];

int lowbit (int x)

{

    return x & -x;

}

void add(int x,int d)

{

    while (x <= n)

    {

        c[x] += d;

        x += lowbit(x);

    }

}

int sum(int x)

{

    int ans = 0;

    while (x > 0)

    {

        ans += c[x];

        x -= lowbit(x);

    }

    return ans;

}

struct Node1

{

    int v,index;

}h[maxn];

struct Node2

{

    int l,r,v,index,ans;

}H[maxn];

bool cmp1(const Node1 &n1,const Node1 &n2)

{

    return n1.v < n2.v;

}

bool cmp2(const Node2 &n1,const Node2 &n2)

{

    return n1.v < n2.v;

}

bool cmp3(const Node2 &n1,const Node2 &n2)

{

    return n1.index < n2.index;

}

int main(void)

{

    #ifndef ONLINE_JUDGE

        freopen("in.txt","r",stdin);

    #endif

    int t,cas = 1;

     scanf ("%d",&t);

    while (t--)

    {

        memset(c,0,sizeof(c));

        scanf ("%d%d",&n,&q);

        for (int i = 1; i <= n; i++)

        {

            scanf ("%d",&h[i].v);

            h[i].index = i;

        }

        sort(h+1,h+n+1,cmp1);

        for (int i = 1; i <= q; i++)

        {

            scanf ("%d%d%d",&H[i].l,&H[i].r,&H[i].v);

            H[i].l++;

            H[i].r++;

            H[i].index = i;

        }

        sort(H+1,H+q+1,cmp2);

        int j = 1;

        for (int i = 1; i <= q; i++)

        {

            int tmp = H[i].v;

            while (h[j].v <= tmp&&j<=n)      //这里要加j<=n 不然会死循环

            {

                add(h[j].index,1);

                j++;

            }

            H[i].ans = sum(H[i].r) - sum(H[i].l-1);

        }

        sort(H+1,H+q+1,cmp3);

        printf("Case %d:\n",cas++);

        for (int i = 1; i <= q; i++)

        {

            printf("%d\n",H[i].ans);

        }

    }

    return 0;

}

View Code