POJ 3067 —— 树状数组求逆序对

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#ACM #树状数组

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Japan

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17885		Accepted: 4825

Description

Japan plans to welcome the ACM ICPC World Finals and a lot of roads must be built for the venue. Japan is tall island with N cities on the East coast and M cities on the West coast (M <= 1000, N <= 1000). K superhighways will be build. Cities on each coast are numbered 1, 2, ... from North to South. Each superhighway is straight line and connects city on the East coast with city of the West coast. The funding for the construction is guaranteed by ACM. A major portion of the sum is determined by the number of crossings between superhighways. At most two superhighways cross at one location. Write a program that calculates the number of the crossings between superhighways.

Input

The input file starts with T - the number of test cases. Each test case starts with three numbers – N, M, K. Each of the next K lines contains two numbers – the numbers of cities connected by the superhighway. The first one is the number of the city on the East coast and second one is the number of the city of the West coast.

Output

For each test case write one line on the standard output:
Test case (case number): (number of crossings)

Sample Input

Sample Output

Test case 1: 5

Source

Southeastern Europe 2006

题意是日本东海岸有m个城市，西海岸有n个城市，修k条公路是两两相连，问你路径的交点数。

可以等效为一个树状数组求逆序对。

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>

using namespace std;
//#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid, r , rt << 1 | 1
//#define mid ((l + r) >> 1)
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 1000 + 50;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
#define mod 1000000007
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pi;
//#pragma comment(linker, "/STACK:102400000,102400000")
int n , m , k;
__int64 c[MAXN];
struct Node
{
    int u , v;
}edge[MAXN * MAXN];
bool cmp(const Node &a , const Node & b)
{
    if(a.u == b.u)return a.v > b.v;
    return a.u > b.u;
}
__int64 lowbit(__int64 x)
{
    return x & (-x);
}
void update(int x)
{
    while(x <= MAXN)
    {
        c[x] += 1;
        x += lowbit(x);
    }
}
__int64 getsum(int x)
{
    __int64 ans = 0;
    while(x > 0)
    {
        ans += c[x];
        x -= lowbit(x);
    }
    return ans;
}
int main()
{
    //ios::sync_with_stdio(false);
    #ifdef Online_Judge
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif // Online_Judge
    int t;
    scanf("%d" , &t);
    FORR(kase , 1 , t)
    {
        scanf("%d%d%d" , &n , &m , &k);
        clr(c, 0);
        FOR(i , 0 , k)scanf("%d%d" ,  &edge[i].u , &edge[i].v);
        sort(edge , edge + k , cmp);
        __int64 ans = 0;
//        outstars
        FOR(i , 0 , k)
        {

            ans += getsum(edge[i].v - 1);
//            outstars
            update(edge[i].v);
        }
        printf("Test case %d: %I64d\n" , kase , ans);
    }
    return 0;
}