A Game
IOI'96 - Day 1
Consider the following two-player game played with a sequence of N positive integers (2 <= N <= 100) laid onto a game board. Player 1 starts the game. The players move alternately by selecting a number from either the left or the right end of the sequence. That number is then deleted from the board, and its value is added to the score of the player who selected it. A player wins if his sum is greater than his opponents.
Write a program that implements the optimal strategy. The optimal strategy yields maximum points when playing against the "best possible" opponent. Your program must further implement an optimal strategy for player 2.
PROGRAM NAME: game1
INPUT FORMAT
| Line 1: |
N, the size of the board |
| Line 2-etc: |
N integers in the range (1..200) that are the contents of the game board, from left to right |
SAMPLE INPUT (file game1.in)
6
4 7 2 9
5 2
OUTPUT FORMAT
Two space-separated integers on a line: the score of Player 1 followed by the score of Player 2.
SAMPLE OUTPUT (file game1.out)
18 11
Analysis
A typical game theory problem. The problem aims to find a solution for two players, which impletement optimal strategy for player 2. Considering the fact of only two operations for each step, making dynamic programming is helpful for these. F[i][j] is defined as the maximum sum for player 1 when choosing numbers among range [i,j], and s[i][j] stands for the sum of numbers among ith number and jth number. After all, new an array num[i] to store numbers.
Then, considering the two activities in choosing, which can only be happened for numbers of the first and the last one. Since the description needs an optimal strategy for player two, it is reasonable for maximum the result.
After all, the function can be presented below: f[i][j]=max{num[i]+sum[i+1][j]-f[i+1][j],num[j]+sum[i][j-1]-f[i][j-1]}, 1<=i,j<=n.
Initialization: f[i][i]=num[i], 1<=i<=n.
Tips in calculating: just only two variables: x,k. "j" is better fixed by x+k for f[i][j] is fully determined by f[i+1][j] and f[i][j-1], which are turing into f[x+1][x+k] and f[x][x+k-1]. It is obvious to making the first "for loop" of k and the next one is "x". Otherwise, i and j are too flexible to make loops.
Code
/**//*
ID:braytay1
PROG:game1
LANG:C++
*/
#include <iostream>
#include <fstream>
using namespace std;
int n;
int f[100][100],num[100],sum,s[100][100];
int max(int a,int b)
{
return a>b?a:b;
}
int main(){
ifstream fin("game1.in");
ofstream fout("game1.out");
fin>>n;
for(int i=0;i<n;i++) fin>>num[i];
memset(f,0,sizeof f);
memset(s,0,sizeof s);
for(int i=0;i<n;i++) f[i][i]=num[i];
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
for(int k=i;k<=j;k++)
s[i][j]+=num[k];
for(int k=1;k<n;k++)
for(int x=0;x+k<n;x++)
f[x][x+k]=max(num[x]+s[x+1][x+k]-f[x+1][x+k],num[x+k]+s[x][x+k-1]-f[x][x+k-1]);
fout<<f[0][n-1]<<" "<<s[0][n-1]-f[0][n-1]<<endl;
return 0;
}