poj 2626 三维dp 0 1 bag

Chess

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2529 Accepted: 1159

Description

The Association of Chess Monsters (ACM) is planning their annual team match up against the rest of the world. The match will be on 30 boards, with 15 players playing white and 15 players playing black. ACM has many players to choose from, and they try to pick the best team they can. The ability of each player for playing white is measured on a scale from 1 to 100 and the same for playing black. During the match a player can play white or black but not both. The value of a team is the total of players' abilities to play white for players designated to play white and players' abilities to play black for players designated to play black. ACM wants to pick up a team with the highest total value.

Input

Input consists of a sequence of lines giving players' abilities. Each line gives the abilities of a single player by two integer numbers separated by a single space. The first number is the player's ability to play white and the second is the player's ability to play black. There will be no less than 30 and no more than 1000 lines on input.

Output

Output a single line containing an integer number giving the value of the best chess team that ACM can assemble.

Sample Input

87 8466 7886 9493 8772 10078 6360 9177 6477 9187 7369 6280 6881 8374 6386 6853 8059 7368 7057 9493 6274 8070 7288 8575 9971 6677 6481 9274 5771 6382 9776 56

Sample Output

2506

题意：

         这里就是说需要找十五个下黑棋，十五个下白棋子的人

 题目中每一个人下  黑  白的不同的战斗力，然后求出这个队可以选出的最大的战斗力

三维dp，以dp[i][j][k]表示从前i个人中选j个下黑子k个下白子的最大能力值，以 a[ i ][ 0 ]和 a[ i ][ 1 ]表示第i名选手下黑白子的能力，那么轻易得出递推方程 

dp[i][j][k]=max(dp[i][j][k],dp[i-1][j-1][k]+a[ i ][ 0 ],dp[i-1][j][k-1]+a[ i ][ 1 ] ) 

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int a[1010][3];int dp[1010][20][20];int main(){    int pos=0;    while(~scanf("%d%d",&a[pos][0],&a[pos][1]))        pos++;    dp[0][1][0]=a[0][0];    dp[0][0][1]=a[0][1];    for(int i=1;i<pos;i++)        for(int j=0;j<16;j++)            for(int k=0;k<16;k++){                if(j)                    dp[i][j][k]=max(dp[i][j][k],max(dp[i-1][j][k],dp[i-1][j-1][k]+a[i][0]));                if(k)                    dp[i][j][k]=max(dp[i][j][k],max(dp[i-1][j][k],dp[i-1][j][k-1]+a[i][1]));            }    printf("%d\n",dp[pos-1][15][15]);    return 0;}