Gentlemen URAL - 1244 [ 01背包 + 记录路径]

1244. Gentlemen

Time limit: 0.5 second
Memory limit: 64 MB

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.

Samples

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

Problem Author: Alexander Petrov
Problem Source: Ural State University Personal Programming Contest, March 1, 2003

Tags:  dynamic programming   ( hide tags for unsolved problems)

Difficulty: 273     Printable version      Submit solution      Discussion (35)

All submissions (11521)    All accepted submissions (2219)    Solutions rating (1564)

题目：给一个v，n个数中找和为v的，如果没有解输出0，如果有多组解输出-1,否则输出删掉那几个数剩下数字的编号。

分析：01背包;

代码：

#include<bits/stdc++.h>
#include<set>
#define ll long long

using namespace std;
const int maxn = 1e2+7;
int a[maxn], dp[maxn*1000], pre[maxn*1000], v, n;
bool vis[maxn];

int main()
{
    //freopen("in.txt", "r", stdin);
    while(~scanf("%d%d", &v, &n))
    {
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        memset(dp, 0, sizeof(dp));
        dp[0] = 1;
        for(int i = 1; i <= n; i++)
        {
            for(int j = v - a[i];j >= 0; j--)
            {
                if(dp[j])
                {
                    if(dp[j + a[i]] == 0&&dp[j] != -1)
                    {
                        dp[j+a[i]] = 1;
                        pre[j+a[i]] = i;
                    } else dp[j+a[i]] = -1;
                }
            }
        }
        memset(vis, 0, sizeof(vis));
        if(dp[v] > 0) {
            int temp = v;
            while(temp)
            {
                vis[pre[temp]] = true;
                temp -= a[pre[temp]];
            }
            int len = 0;
            for(int i = 1; i <= n; i++)
            {
                if(!vis[i])
                {
                    if(len == 0) printf("%d", i);
                    else printf(" %d", i);
                    len++;
                }
            }
            printf("\n");
        } else printf("%d\n", dp[v]);
    }
    return 0;
}