POJ3067 树状数组 情景转化为区间求和

0）

题目大意：

小岛上东、西各有一列从北到南依次排列的城市，现在将东边与西边的指定城市之间建造高速公路（不同列之间的城市相连），要求给出交叉点个数（假设没有大于两条以上的道路会交于一点）。

转化：

随着连接的线段越来越多，交点个数应该只增不降，而输入数据给出的顺序已经是一个提示，我们假设第一列城市属于l列，第二列城市属于r列，那么l列输入的顺序是从小到大时，可以保证l列当前城市（假设是城市2）连接到的r列某城市（假设是城市3），一定是在l列当前城市之前的城市（假设是城市1）连接到的r列城市之前（假设是4）才会构成交点。因此，我们只需不断返回连接当前线段的同时又加了几个交点即可。而在l列城市顺序输入（编号从小到大）时，而所加交点个数，正对应连接当前线段时被连接到的r列城市（城市3）到最后一个城市（城市6）之间所有城市的（不包括城市3，所以恰好sum(6)-sum(3)，而不用sum(6)-sum(3-1)）被连接次数之和。于是问题，似乎变成了求任意区间求和，线段树和树状数组都可以解决。

注意：如果题目中给出的城市编号有0，就像下面这张图，那么我们应当将所有城市编号++，再用树状数组或线段树。





1）

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;
const int maxn=1010;
struct Road{
int l;
int r;
}road[maxn*maxn];//注意，道路是城市个数的乘积,不然Runtime
int c[maxn];
bool cmp(Road a,Road b){
if(a.l<b.l){
return 1;
}
else if(a.l==b.l){
return a.r<=b.r;//cout<<a>=b>>endl;如果相等输出什么?
}
return 0;
}
int Sum(int x){
long long int sum=0;
while(x>0){
sum+=c[x];
x-=(x&(-x));
}
return sum;
}
void Add(int x){
while(x<maxn){
c[x]++;
x+=(x&(-x));
}
}
int main()
{
int t;
scanf("%d",&t);
int kase=0;
while(t--){
kase++;
memset(c,0,sizeof(c));
int n,m,k;
scanf("%d%d%d",&n,&m,&k);
for(int i=0;i<k;i++){
scanf("%d%d",&road[i].l,&road[i].r);
}
sort(road,road+k,cmp);
long long int res=0;
for(int i=0;i<k;i++){
res+=Sum(m)-Sum(road[i].r);
Add(road[i].r);
}
printf("Test case %d: %lld\n",kase,res);
}
}

2）

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