URAL1224——背包DP+回溯——Gentlemen

Description

Let's remember one old joke:

Once a gentleman said to another gentleman:
— What if we play cards?
— You know, I haven't played cards for ten years…
— And I haven't played for fifteen years…
So, little by little, they decided to resurrect their youth. The first gentleman asked a servant to bring a pack of cards, and before starting playing out weighed in his hand the pack.
— It seems to me, one card is missing from the pack… — he said and gave the pack to the other gentleman.
— Yes, the nine of spades, — the man agreed.

An incomplete pack of cards is given. The program should determine which cards are missing.

Input

The first line contains a positive integer, which is the weight in milligrams of the given incomplete pack. The second line contains an integer  N, 2 ≤  N ≤ 100 — the number of cards in the complete pack. In the next  N lines there are integers from 1 to 1000, which are the weights of the cards in milligrams. It's guaranteed that the total weight of all cards in the complete pack is strictly greater than the weight of the incomplete pack.

Output

If there is no solution, then output the single number 0. If there are more than one solutions, then you should write −1. Finally, if it is possible to determine unambiguously which cards are missing in the incomplete pack as compared to the complete one, then output the numbers of the missing cards separated with a space in ascending order.

Sample Input

input	output
270 4 100 110 170 200	2 4
270 4 100 110 160 170	-1
270 4 100 120 160 180	0

大意：输入一个m,表示当前剩下的纸牌的重量，接下来n行表示完整的牌的重量，每种牌只有一张，让你求缺少的牌是哪些。

用所有的牌的总重量减去当前不完整牌的重量，就是所缺少的牌的重量，然后进行01背包，不过dp记录的是情况数目所以回溯的时候不能用i = path[i]，只能枚举倒着递推

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

int w[110];

int dp[100100];

int path[100100];

int print[110];

int main()

{

    int m,n;

    while(~scanf("%d%d",&m,&n)){

        int sum = 0;

        for(int i = 1; i <= n ; i++){

            scanf("%d",&w[i]);

            sum += w[i];

        }

        memset(dp,0,sizeof(dp));

        dp[0] = 1;

        sum -= m;

       for(int i = 1; i <= n ;i++){

           for(int j = sum - w[i]; j >= 0 ;j--){

               if(dp[j]){

                    if(!dp[j+w[i]])

                        path[j+w[i]] = i;

                    dp[j+w[i]] += dp[j];

               }

           }

       }

       if(dp[sum] == 0)

           printf("0\n");

       else if(dp[sum] == 1){

           int j = 1;

           for(int i = n ; i >= 1; i --){

               if(path[sum] == i){

                   print[j++] = i;

                   sum -= w[i];

               }

           }

           for(int i = j - 1; i > 1;i--)

               printf("%d ",print[i]);

           printf("%d\n",print[1]);

       }

       else printf("-1\n");

    }

    return 0;

}