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

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

Give you n groups of numbers, each group has m, two people the first person can only get from the beginning, the second person can only get from the last one, the first person first hand, everyone is smart enough to ask you what their maximum score is.

Solution: Obviously, if each group can obtain symmetric numbers, the total score will not be affected. The number of groups with an odd number will be affected, and their priority will be related to the first hand, first, remove the largest one, and then remove the largest one, and so on ..

C. Fox and Card Gametime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

Fox Ciel is playing a card game with her friend Fox Jiro. There areNPiles 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 integerN(1 digit ≤ DigitNLimit ≤0000100). Each of the nextNLines contains a description of the pile: the first integer in the line isSI(1 digit ≤ DigitSILimit ≤ limit 100)-the number of cards inI-Th pile; then followSIPositive integersC1,C2 ,...,CK,...,CSI(1 digit ≤ DigitCKLimit ≤ limit 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

21 1002 1 10

Output

101 10

Input

19 2 8 6 5 9 4 7 1 3

Output

30 15

Input

33 1 3 23 5 4 62 8 7

Output

18 18

Input

33 1000 1000 10006 1000 1000 1000 1000 1000 10005 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;    }}