POJ1837 DP

POJ 1837 DP題
タイトルは最初にN久を见ました...意味は大体1つの天秤があって、左の腕の长さは-15から0で、右の腕の长さは0から15で、あなたにcつのフックをあげて、gの分銅、すべての分銅の重さはすべて1から25で、すべての分銅を天秤の上に挂けてそしてバランスを取る方案はいくつありますかを闻きます.バランスをとるには物理的知識でモーメント=0、左重量X左アーム長+右重量X右アーム長=0となるため、状態は全部で25*15*20=7500となり、dp[i][j]を前のi個の分銅をバランスポイントjに掛ける案とするとともに、この問題に負数があるので座標軸を0~15000に移動し、dp[0][7500]=1を初期化する
Description
Gigel has a strange “balance” and he wants to poise it. Actually, the device is different from any other ordinary balance. It orders two arms of negligible weight and each arm’s length is 15. Some hooks are attached to these arms and Gigel wants to hang up some weights from his collection of G weights (1 <= G <= 20) knowing that these weights have distinct values in the range 1..25. Gigel may droop any weight of any hook but he is forced to use all the weights. Finally, Gigel managed to balance the device using the experience he gained at the National Olympiad in Informatics. Now he would like to know in how many ways the device can be balanced.
Knowing the repartition of the hooks and the set of the weights write a program that calculates the number of possibilities to balance the device. It is guaranteed that will exist at least one solution for each test case at the evaluation.
Input
The input has the following structure: • the first line contains the number C (2 <= C <= 20) and the number G (2 <= G <= 20); • the next line contains C integer numbers (these numbers are also distinct and sorted in ascending order) in the range -15..15 representing the repartition of the hooks; each number represents the position relative to the center of the balance on the X axis (when no weights are attached the device is balanced and lined up to the X axis; the absolute value of the distances represents the distance between the hook and the balance center and the sign of the numbers determines the arm of the balance to which the hook is attached: ‘-’ for the left arm and ‘+’ for the right arm); • on the next line there are G natural, distinct and sorted in ascending order numbers in the range 1..25 representing the weights’ values.
Output
The output contains the number M representing the number of possibilities to poise the balance.
Sample Input
2 4 -2 3 3 4 5 8
Sample Output
2
コード#コード#

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;

int dp[21][15002];

int main()
{
    int c,g;
    while (scanf("%d%d",&c,&g)!=EOF)
    {
        int post[21];
        int weight[21];
        for (int i =1 ;i<=c;i++)
           scanf("%d",&post[i]);
        for (int i = 1;i<=g;i++)
            scanf("%d",&weight[i]);
        memset(dp,0,sizeof(dp));
        dp[0][7500] = 1;                //      ，       

        for (int i = 1;i<=g;i++)
          for (int j = 0;j<=15000;j++)
            {
             for (int k = 1;k<=c;k++)
              if (j >= post[k] * weight[i])
                 dp[i][j]+=dp[i-1][j-post[k]*weight[i]];
            }
        printf("%d
",dp[g][7500]);
    }
}

上のプログラムは94 MS走って、最適化を試みて47 MSになって、実はただ2つの循環が位置を変えただけです

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;

int dp[21][15002];

int main()
{
    int c,g;
    while (scanf("%d%d",&c,&g)!=EOF)
    {
        int post[21];
        int weight[21];
        for (int i =1 ;i<=c;i++)
           scanf("%d",&post[i]);
        for (int i = 1;i<=g;i++)
            scanf("%d",&weight[i]);
        memset(dp,0,sizeof(dp));
        dp[0][7500] = 1;                //      ，       

        for (int i = 1;i<=g;i++)
          for (int k = 1;k<=c;k++)
            for (int j = 15000;j>=post[k]*weight[i];j--)
            {
                 dp[i][j]+=dp[i-1][j-post[k]*weight[i]];
            }
        printf("%d
",dp[g][7500]);
    }
}

大牛の..0 MS走った....

#include <iostream>
using namespace std;

int dp[21][20000];

int main(){
    int i,j,k,C,G;

    int pos[21],wight[21];
    while(scanf("%d%d",&C,&G)!=EOF){
        for(i=1;i<=C;i++)
            scanf("%d",&pos[i]);
        for(i=1;i<=G;i++)
            scanf("%d",&wight[i]);

        memset(dp,0,sizeof(dp));
        dp[0][10000]=1;

        for(i=1;i<=G;i++)
            for(j=0;j<=20000;j++)
                if(dp[i-1][j]){
                    for(k=1;k<=C;k++)
                        dp[i][j+pos[k]*wight[i]]+=dp[i-1][j];
                }

        printf("%d
",dp[G][10000]);
    }
    return 0;
}