POJ3067 Japan(树状数组,逆序数)

题目：

Japan

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 27465 Accepted: 7428

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

13 4 41 42 33 23 1

Sample Output

Test case 1: 5

Source

Southeastern Europe 2006

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思路：

题意：日本岛东海岸与西海岸分别有N和M个城市，现在修高速公路连接东西海岸的城市，求交点个数。
做法：记每条告诉公路为(x,y), 即东岸的第x个城市与西岸的第y个城市修一条路。当两条路有交点时，满足（x1-x2）*（y1-y2） < 0。所以，将每条路按x从小到达排序，若x相同，按y从小到大排序。 然后按排序后的公路用树状数组在线更新，求y的逆序数之 和 即为交点个数。
上面说的可能有点难理解，详细说明如下。
记第i条边的端点分别为xi,yi。
由于x是从小到大排序的，假设当前我们在处理第k条边，那么第1~k-1条边的x必然是小于（等于时候暂且不讨论）第k条边的 x 的，那么前k-1条边中，与第k条边相交的边的y值必然大于yk的，所以此时我们只需要求出在前k-1条边中有多少条边的y值在区间[yk, M]即可,也就是求yk的逆序数，M为西岸城市个数，即y的最大值。 所以就将问题转化成区间求和的问题，树状数组解决。当两条边的x相同时，我们记这两条边的y值分别为ya，yb（ya<yb),我们先处理（x，ya），再处理（x，yb），原因很明显，因为当x相同时，这两条边是认为没有交点的，若先处理（x，yb），那么下次处理（x，ya）时，（x，ya）就会给（x，yb）增加一个逆序，也就是将这两条边做相交处理了。

代码：

#include <cstdio>#include <cstring>#include <cctype>#include <string>#include <set>#include <iostream>#include <stack>#include <cmath>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define mod 10000007#define debug() puts("what the fuck!!!")#define N (1000+20)#define ll long longusing namespace std;ll c[N];struct node{    ll x,y;} g[N*N];bool cmp(node a,node b){    if(a.x==b.x)        return a.y<b.y;    return a.x<b.x;}ll lowbit(ll x)//用来维护树状数组,得到管辖区间{    return x&-x;}void add(ll k,ll num,ll n)//给数组的第k位增加num个{    while(k<=n)    {        c[k]+=num;        k+=lowbit(k);    }}ll sum(ll n)//求要查询的区间的前n项和{    ll sum=0;    while(n>0)    {        sum+=c[n];        n-=lowbit(n);    }    return sum;}int main(){    ll t,n,m,k,q=1;    scanf("%lld",&t);    while(t--)    {        mem(c,0);        scanf("%lld%lld%lld",&n,&m,&k);        for(ll i=1; i<=k; i++)            scanf("%lld%lld",&g[i].x,&g[i].y);        sort(g+1,g+k+1,cmp);        ll ans=0;        for(ll i=1; i<=k; i++)        {            add(g[i].y,1,m);            ans+=(sum(m)-sum(g[i].y));        }        printf("Test case %lld: %lld\n",q++,ans);    }    return 0;}

参考博客：http://blog.csdn.net/weiguang_123/article/details/7895848