D - Fox and Card Game
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Description
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 Input
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
Hint
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.
首先題意就是兩個人在放了許多堆牌的桌面拿牌，每張牌都有每張牌的值，並且Ciel 只能從牌堆頂上拿牌，Jiro 只能從牌堆的底部拿牌，問如果兩個人拿牌得當的話，最終每人拿到的牌的值得總和是多少。
很明顯我們要分析對於偶數堆牌，如果牌堆的上半部分比下半部分總和大，那Jiro肯定不會浪費次數來拿拿不到的牌 ，肯定會去搶奇數堆牌的中間大的那堆，同時Ciel 也知道，對面如果要搶偶數堆牌的上半部分，自己只要跟着他那就行了，這樣對面就拿不到了，所以，每個人要搶的其實只有奇數堆牌中間的那一張牌，因爲除去那一張牌後，奇數堆牌會變得和偶數堆牌一樣對半分。
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
int st[105];
bool cmp(int a,int b){
return a > b;}
int main()
{
int n,x,y;
int l = 0,r = 0,cnt = 0;
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d",&x);
int tmp ;
if(x&1) tmp = (x+1)/2;
else tmp = x/2;
for(int j=1;j<=x;j++){
scanf("%d",&y);
if(j<=tmp) l+=y;
else r+= y;
if(x&1 && j==tmp){
st[cnt++] = y;
}
}
if(x&1) l -= st[cnt-1];
}
sort(st,st+cnt,cmp);
for(int i=1;i<=cnt;i++){
if(i&1) l += st[i-1];
else r += st[i-1];
}
printf("%d %d\n",l,r);
return 0;
}