（POJ 3067） Japan （慢慢熟悉的树状数组）

Japan

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

1
3 4 4
1 4
2 3
3 2
3 1

Sample Output

Test case 1: 5

看似是一道二分图图论的问题，其实经过一些简单的（大雾）变换之后，就可以转化为树状数组来求解自昨天被辉大神教做人之后，我对树状数组的理解也有了一些加深：0

大体题意：日本有东和西两个岛，从北到南依次标号为1,2,3...然后给出k条边，求出k条边一共有多少交点（注意是两两相交，只要两条线有交点就算一个）。

经过分析之后可以发现如果两条边有交点，就必须满足x1-x2>0&&y1-y2<0或者x1-x2<0&&y1-y2>0;有两个条件，所以我们可以通过将x排序来变为一个条件：即将所有边按x的大小从小到大排序，然后一边遍历，一边维护树状数组。对于边k，在边1到边k-1中寻找y>yk的边，有几条这样的边，那么对于边k，在1~k-1中就有几个与之相交的边。一边维护树状数组，一边更新交点的值，遍历结束后就可求出所有交点的数量。

PS：在排序时，如果x1==x2,那么要按y1<y2的顺序来排，不然在计算与较小y值那条边相交边的数量时，会将x与之相等的边也算进去（因为树状数组只存储y的值）。

#include<iostream>
#include<cstdio>
#include<vector>
#include<set>
#include<map>
#include<string.h>
#include<cmath>
#include<algorithm>
#include<queue>
#define LL long long
#define inf 0x3f3f3f3f

using namespace std;
int n,m,k;

struct de
{
    int x,y;
    bool operator<(const de &other)const
    {
        if(x==other.x)
            return y<other.y;
        return x<other.x;
    }
};

de qiao[500*1010];
LL run[10100]={0};

int lowbit(int i){return i&(-i);}

void update(int i)
{
    while(i<=m)
    {
        run[i]++;
        i+=lowbit(i);
    }
}

LL sum(int i)
{
    LL s=0;
    while(i>0)
    {
        s+=run[i];
        i-=lowbit(i);
    }
    return s;
}

int main()
{
    int t;
    scanf("%d",&t);
    for(int p=1;p<=t;p++)
    {
        scanf("%d%d%d",&n,&m,&k);
        memset(qiao,0,sizeof(qiao));
        memset(run,0,sizeof(run));
        for(int i=1;i<=k;i++)
            scanf("%d%d",&qiao[i].x,&qiao[i].y);
        sort(qiao+1,qiao+1+k);
        LL ans=0;
        for(int i=1;i<=k;i++)
        {
            ans+=sum(m)-sum(qiao[i].y);
            update(qiao[i].y);
        //cout << "ans=" << ans << endl;

        }
        printf("Test case %d: %I64d\n",p,ans);
    }
    return 0;
}