zjnu1725 COCI (类似二维树状数组模拟)

Description

The 3rd round of COCI is already here! In order to bet on predict the scores, we have assumed the following:

If contestant A scored strictly more points than contestant B in each of the first two rounds, then in the

third round A will score at least an equal amount of points as B.

Of course, in each round (including this one, the 3rd one) it is possible to score from 0 to 650 points. On the

total ranking list, contestants are sorted descending according to the sum of points from all three rounds.

The contestants with an equal sum share the same place and the next contestant gets the realistic following place.

For example, contestants with sums equal to 1000, 1000, 900, 900 and 800 points win places 1., 1., 3., 3. and 5.,

respectively.

For each of the N contestants, we know the number of points scored in the first and second round. Given the

aforementioned assumption, determine the highest and lowest place each contestant can get on the total ranking

list after three rounds of COCI.

Input

The first line of input contains an integer N (1 <= N <= 500000), the number of contestants.

Each of the following N lines contains two integers from the interval [0; 650]: the number of points each contestant

won in the first and second round.

Output

For each contestant, in the order given in the input, output two integers per line: the required highest and lowest

place they can get on the total ranking list.

Sample Input

5

250 180

250 132

220 123

132 194

220 105

10

650 550

550 554

560 512

610 460

610 456

650 392

580 436

650 366

520 456

490 456

Sample Output

1 3

1 3

3 5

1 5

3 5

1 4

1 8

2 8

2 7

2 9

1 10

4 10

1 10

5 10

5 10

题意：n个人进行决斗，共有三场比赛，前两场比赛的成绩已经知道了，第三场的成绩未知，已知如果A前两场比赛每一场比赛成绩都严格高于B，那么A的第三场成绩一定大于等于B，最后的结果按三场总成绩从高到低排序，问每一个人最好的名次和最差的名词分别可能为多少。

思路：先考虑一个人A最好的情况，那么A得到的分数为650分，那么之前两场严格高于她的人最后一定也高于她，如果B之前一门小于等于她，另一门高于她，这种情况最后也不会比她高，因为可以另B的第三门成绩为0，那么A的成绩一定是大于等于B的（等于的情况有两种，一种是第二门成绩为0，但是B的第二门成绩为650，第一门成绩两者相同，另一种是第一种第二种反一下），又因为两者成绩相等自己不会退后名词，所以可以不用管。再考虑A最坏的情况，我们考虑比A考的差的人，这里和之前一样，只是要考虑(0,0,a[i]-1,b[i]-1)的总数，这里还要考虑和她成绩相同的情况，就是A的a[i]或者b[i]=650,要减去这些，然后用n减去比她考的差的人数就是答案了。

注意：这题不能用树状数组，因为会超时，又因为点的坐标大小都很小，所以可以用s[i][j]表示x小于等于i，y小于等于j的坐标总数。

#include<iostream>

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

#include<math.h>

#include<vector>

#include<map>

#include<set>

#include<queue>

#include<stack>

#include<string>

#include<bitset>

#include<algorithm>

using namespace std;

typedef long long ll;

typedef long double ldb;

#define inf 99999999

#define pi acos(-1.0)

#define NN 500050

#define MM 650

int a[NN],b[NN];

int k[MM+10][MM+10],s[MM+10][MM+10];

int solve(int p,int q,int P,int Q)

{

    int i,j;

    if(P<0 || Q<0)return 0;

    int sum=s[P][Q];

    if(p>0){

        sum-=s[p-1][Q];

    }

    if(q>0){

        sum-=s[P][q-1];

    }

    if(p>0 && q>0){

        sum+=s[p-1][q-1];

    }

    return sum;

}

int main()

{

    int n,m,i,j,c,d;

    while(scanf("%d",&n)!=EOF)

    {

        memset(k,0,sizeof(k));

        memset(s,0,sizeof(s));

        for(i=1;i<=n;i++){

            scanf("%d%d",&a[i],&b[i]);

            k[a[i] ][b[i] ]++;

        }

        for(i=0;i<=MM;i++){

            for(j=0;j<=MM;j++){

                s[i][j]=k[i][j];

                if(i>0){

                    s[i][j]+=s[i-1][j];

                }

                if(j>0){

                    s[i][j]+=s[i][j-1];

                }

                if(i>0 && j>0){

                    s[i][j]-=s[i-1][j-1];

                }

            }

        }

        for(i=1;i<=n;i++){

            printf("%d ",solve(a[i]+1,b[i]+1,MM,MM)+1 );

            printf("%d\n",n-solve(0,0,a[i]-1,b[i]-1)-k[a[i] ][0]*(b[i]==MM)-k[0][b[i] ]*(a[i]==MM)   );

        }

    }

    return 0;

}

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zjnu1725 COCI (类似二维树状数组模拟)

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