TZOJ 5101 A Game(區間DP)

描述

Consider the following two-player game played with a sequence of N positive integers (2 <= N <= 100) laid onto a 1 x N 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 gameboar. 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.

輸入

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 ??

輸出

Two space-separated integers on a line: the score of Player 1 followed by the score of Player 2.

樣例輸入

6
4 7 2 9
5 2

樣例輸出

18 11

題意

1*N的游戲盤，每個格子都有價值，玩家一次拿最左或最右，拿了刪掉這個格子，問玩家1先拿，玩家2后拿，問倆人都最優可以拿多少分

題解

dp[i][j]=區間[i,j]先手-后手的差值

很容易推出dp[i][j]可由小區間得到

dp[i][j]=a[i]-dp[i+1][j];

dp[i][j]=a[j]-dp[i][j-1];

最終我們得到dp[1][n]為先手-后手的差值

設玩家1拿了V1,玩家2拿了V2

V1+V2=Σa;

V1-V2=dp[1][n];

求得

V1=(Σa+dp[1][n])/2;

V2=(Σa-dp[1][n])/2;

代碼

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int a[105],dp[105][105]={0},sum=0;
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]),sum+=a[i];
        for(int len=1;len<=n;len++)
            for(int i=1;i<=n-len+1;i++)
            {
                int j=i+len-1;
                dp[i][j]=max(a[i]-dp[i+1][j],a[j]-dp[i][j-1]);
            }
        printf("%d %d\n",(sum+dp[1][n])/2,(sum-dp[1][n])/2);
    }
    return 0;
}