3067 Japan 树状数组

Japan

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8532		Accepted: 2290

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

Source

Southeastern Europe 2006

WA了差不多10次。。。。。悲剧啊。。。。虽然N,M是到1000，可是节点的数量是会超INT的范围的。。。。接着就是节点是否重合的问题。。。。这个真的很难办。。。。最后我直接将Y俺减序排列，接着X按减序排，最后记录下每个X重合的数目，因为X的范围是1000，接着得到SUM,再减去每个X重合的数目Y  (Y-1)Y/2，这一条是画图看出来的

#include<stdio.h>

#include<algorithm>

#include<string.h>

using namespace std;

struct t

{

    int x,y;

    __int64 value;

}e[1000010];

int n,m,k;

__int64 tree[1005];

bool cmp(t a, t b)

{

    if(a.y!=b.y)  return a.y>b.y;

    return a.x>b.x;

}

inline int Lowbit(int x)

{

    return x&(-x);

}

void Update(int x)

{

    for(int i=x;i<=1000;i+=Lowbit(i))

    tree[i]++;

}

__int64 Getsum(int x)

{

    __int64 temp=0;

    for(int i=x;i>0;i-=Lowbit(i))

         temp+=tree[i];

    return temp;

}

int main()

{

    int T,cas=1;

    scanf("%d",&T);

    while(T--)

    {

        memset(tree,0,sizeof(tree));

        scanf("%d%d%d",&n,&m,&k);

        for(int i=1;i<=k;i++)

           scanf("%d%d",&e[i].x,&e[i].y);

        sort(e+1,e+k+1,cmp);

        for(int i=1;i<=k;i++)

        {

            e[i].value=Getsum(e[i].x);

            Update(e[i].x);

        }

        int d=0;

        __int64 sum=0;

        int ss[1001]={0};

        for(int i=1;i<=k;i++)

        {

            ss[e[i].x]++;

            sum+=e[i].value;

         }

         for(int i=1;i<1001;i++)

         {

             sum-=(ss[i])*(ss[i]-1)/2;

         }

        printf("Test case %d: %I64d/n",cas++,sum);

    }

    return 0;

}