POJ 2481 Cows

题意：有n头，他们都有两个属性。第i头牛的属性为s[i]和e[i] (s[i] < e[i])，定义它比第j头牛强壮即是s[i] <= s[j] && e[i] >= e[j] && e[i]-s[i] != e[j]-s[j]。对每投牛求比它强壮的牛有多少头。

　　n <= 10^5, s[i],e[i] <= 10^5

解法：首先，对牛排序，按e[i]从大到小，e[i]相同则s[i]从小到大。则扫描到某头牛i的时候，比它强壮的牛都在它的前面(因为s[i]==s[j]&&e[i]==e[j]不算比它强壮)而且一定有它前面的牛e值大等于e[i]，所以，只需要求s[i]小等于它的有多少个(如果牛i与牛i-1属性相同，则所求数量也相同，不用再计算)。

　　　考虑到s[i] <= 10^5，所以直接用num[10^5]记录即可，然后求前缀和。由于求前缀和的同时，也需要动态的修改，所以用树状数组。

Ps:感觉这道题也可以用二分来做，只不过在做专题，所以还是用树状数组做了。

tag:BIT

/*
 * Author:  Plumrain
 * Created Time:  2014-02-16 21:05
 * File Name: BIT-POJ-2481.cpp
 */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#include <vector>
#include <cstdlib>
#include <sstream>
#include <fstream>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cctype>
#include <ctime>
#include <utility>

using namespace std;

#define clr(x) memset(x, 0, sizeof(x))
#define clrs(x,y) memset(x, y, sizeof(x))
#define pb push_back
#define sz(v) ((int)(v).size())
#define all(t) t.begin(),t.end()
#define INF 999999999999999999
#define zero(x) (((x)>0?(x):-(x))<eps)
#define out(x) cout<<#x<<":"<<(x)<<endl
#define tst(a) cout<<a<<" "
#define tst1(a) cout<<#a<<endl
#define lowbit(x) ((x)&(-x))
#define CINBEQUICKER std::ios::sync_with_stdio(false)

const double eps = 1e-8;
const double PI = atan(1.0)*4;
const int inf = 2147483647 / 2;

typedef long long int64;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<double> vd;
typedef pair<int, int> pii;

inline int Mymod (int a, int b) {int x=a%b; if(x<0) x+=b; return x;}

struct nod{
    int s, e, idx;
};

nod an[100005];
int num[100005], c[100005], lim;
vector<pii > ans;

int sum(int x)
{
    int ret = 0;
    while (x > 0)
        ret += c[x], x -= lowbit(x);
    return ret;
}

void add(int x, int d)
{
    while (x <= lim)
        c[x] += d, x += lowbit(x);
}

bool cmp1(nod a, nod b)
{
    return a.e > b.e || (a.e == b.e && a.s < b.s);
}

bool cmp2(pii a, pii b)
{
    return a.second < b.second;
}

int main()
{
//    freopen("a.in","r",stdin);
//    freopen("a.out","w",stdout);
//    std::ios::sync_with_stdio(false);
    int n;
    while (scanf ("%d", &n) != EOF && n){
        for (int i = 0; i < n; ++ i){
            scanf ("%d%d", &an[i].s, &an[i].e);
            ++ an[i].s; ++ an[i].e;
            an[i].idx = i;
            lim = max(lim, an[i].s);
        }
        sort (an, an + n, cmp1);

        //for (int i = 0; i < n; ++ i) tst (an[i].s), tst (an[i].e), out (an[i].idx);
//
        clr (num); clr (c);
        ans.clear();
        int cnt = 0;
        for (int i = 0; i < n; ++ i){
            if (!i || an[i].s != an[i-1].s || an[i].e != an[i-1].e)
                cnt = sum(an[i].s);
                
            ans.pb (make_pair(cnt, an[i].idx));
            add (an[i].s, 1);
        }
        sort (all(ans), cmp2);
        
        for (int i = 0; i < sz(ans); ++ i){
            if (i) printf (" ");
            printf ("%d", ans[i].first);
        }
        printf ("
");
    }
    return 0;
}