Codeforces Round #228 (Div. 1) C. Fox and Card Game

题目大意：给你n组数，每组有m个，有两个人第一个人只能从开头开始取，第二个人只能从最后一个取，第一个人先手，每个人都足够聪明，问你他们最后的最大的得分为多少。

解题思路：显然如果每组都可以对称着取数，不会影响总得分，影响的分的是个数为奇数的那几组，他们大小的先后就和先手有关，先手取掉最大的，后手取掉次大的，依次类推。。

C. Fox and Card Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Fox Ciel is playing a card game with her friend Fox Jiro. There are n piles of cards on the table. And there is a positive integer on each card.

The players take turns and Ciel takes the first turn. In Ciel's turn she takes a card from the top of any non-empty pile, and in Jiro's turn he takes a card from the bottom of any non-empty pile. Each player wants to maximize the total sum of the cards he took. The game ends when all piles become empty.

Suppose Ciel and Jiro play optimally, what is the score of the game?

Input

The first line contain an integer n (1 ≤ n ≤ 100). Each of the next n lines contains a description of the pile: the first integer in the line is si(1 ≤ si ≤ 100) — the number of cards in the i-th pile; then follow si positive integers c1, c2, ..., ck, ..., csi (1 ≤ ck ≤ 1000) — the sequence of the numbers on the cards listed from top of the current pile to bottom of the pile.

Output

Print two integers: the sum of Ciel's cards and the sum of Jiro's cards if they play optimally.

Sample test(s)

input

2
1 100
2 1 10

output

101 10

input

1
9 2 8 6 5 9 4 7 1 3

output

30 15

input

3
3 1 3 2
3 5 4 6
2 8 7

output

18 18

input

3
3 1000 1000 1000
6 1000 1000 1000 1000 1000 1000
5 1000 1000 1000 1000 1000

output

7000 7000

Note

In the first example, Ciel will take the cards with number 100 and 1, Jiro will take the card with number 10.

In the second example, Ciel will take cards with numbers 2, 8, 6, 5, 9 and Jiro will take cards with numbers 4, 7, 1, 3.

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define mod 1000000007

#define Read() freopen("autocomplete.in","r",stdin)
#define Write() freopen("autocomplete.out","w",stdout)
#define Cin() ios::sync_with_stdio(false)

using namespace std;

const int maxn = 110;

int mp[maxn][maxn];
int num[maxn];

int main()
{
    int n;
    while(cin >>n)
    {
        int lsum = 0;
        int rsum = 0;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&mp[i][0]);
            for(int j = 1; j <= mp[i][0]; j++)
                scanf("%d",&mp[i][j]);
        }
        int ans = 0;
        for(int i = 1; i <= n; i++)
        {
            int x = mp[i][0]/2;
            ///cout<<"x == "<<x<<endl;
            for(int j = 1; j <= x; j++) lsum += mp[i][j];
            if(mp[i][0]%2)
            {
                num[ans++] = mp[i][x+1];
                x++;
            }
            for(int j = x+1; j <= mp[i][0]; j++) rsum += mp[i][j];
        }
        sort(num, num+ans);
        int cnt = 0;
        for(int i = ans-1; i >= 0; i--)
        {
            if(cnt%2) rsum += num[i];
            else lsum += num[i];
            cnt++;
        }
        cout<<lsum<<" "<<rsum<<endl;
    }
}