HOJ 2091 Chess（三维简单DP）

Chess

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Source : Univ. of Alberta Local Contest 1999.10.16

Time limit : 1 sec Memory limit : 32 M

Submitted : 244, Accepted : 100

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.

There are multiple test cases. Each case will be followed by a single line containing a “*”.

Output

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

Sample Input

87 84

66 78

86 94

93 87

72 100

78 63

60 91

77 64

77 91

87 73

69 62

80 68

81 83

74 63

86 68

53 80

59 73

68 70

57 94

93 62

74 80

70 72

88 85

75 99

71 66

77 64

81 92

74 57

71 63

82 97

76 56

*

Sample Output

2506

如果能想到用三维表示状态，就很容易了

dp[i][j][k]表示到第i个人已经选了j个人去打白选了k个人去打黑

那么dp[i][j][k]只有三种情况，在第i个人要么不选他，要么选他打白，要么选他打黑

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>

using namespace std;
int dp[1005][20][20];
int a[1005];
int b[1005];
char c[1005];
char d[1005];
int fun(char *a)
{
    int len=strlen(a);
    int num=0;
    for(int i=0;i<len;i++)
    {
        num+=(a[i]-'0')*(int)(pow(10,(len-i-1)));
    }
    return num;

}
int main()
{
    while(scanf("%s",c)!=EOF)
    {
        scanf("%s",d);
        int cnt=0;
        while(true)
        {
            a[++cnt]=fun(c);
            b[cnt]=fun(d);
            scanf("%s",c);
            if(c[0]=='*')
                break;
            else
                scanf("%s",d);
        }
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=cnt;i++)
        {
            for(int j=0;j<=15;j++)
            {
                for(int k=0;k<=15;k++)
                {
                    if(j+k>i)
                        continue;
                    if(!j&&k)
                        dp[i][j][k]=max(dp[i-1][j][k],dp[i-1][j][k-1]+b[i]);
                    else if(j&&!k)
                        dp[i][j][k]=max(dp[i-1][j][k],dp[i-1][j-1][k]+a[i]);
                    else if(!j&&!k)
                        dp[i][j][k]=dp[i-1][j][k];
                    else
                        dp[i][j][k]=max(dp[i-1][j][k],max(dp[i-1][j-1][k]+a[i],dp[i-1][j][k-1]+b[i]));
                }
            }
        }
        printf("%d\n",dp[cnt][15][15]);
    }
    return 0;
}