1068 Find More Coins (30 分)(01背包+最优解路径选择)

题目描述:

1068 Find More Coins (30 分)

 

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she must pay the exact amount. Since she has as many as 1 coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find some coins to pay for it.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (≤, the total number of coins) and M (≤, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the face values V​1​​≤V​2​​≤⋯≤V​k​​ such that V​1​​+V​2​​+⋯+V​k​​=M. All the numbers must be separated by a space, and there must be no extra space at the end of the line. If such a solution is not unique, output the smallest sequence. If there is no solution, output "No Solution" instead.

Note: sequence {A[1], A[2], ...} is said to be "smaller" than sequence {B[1], B[2], ...} if there exists k≥1 such that A[i]=B[i] for all i<k, and A[k] < B[k].

Sample Input 1:

8 9
5 9 8 7 2 3 4 1

 

Sample Output 1:

1 3 5

 

Sample Input 2:

4 8
7 2 4 3

 

Sample Output 2:

No Solution
思路：一开始就想到排序+01背包,但是不知道怎么还原最优解路径,然后写了个dfs果然超时了,看了题解才知道应该用二维数组保存每一个物品在不同容量下的可取性,
然后从小到达逆推出可行的结果,这就求得了最优解,排序时要注意从大到小拍,因为我们要从小到大开始选择
AC代码:

#include<iostream>
#include<string.h>
#include<cmath>
#include<set>
#include<map>
#include<string>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>
#include<algorithm>
#include<climits>
using namespace std;
typedef long long ll;
inline int read() {
    int sum = 0, f = 1;
    char p = getchar();
    for (; !isdigit(p); p = getchar()) if (p == '-')f = -1;
    for (; isdigit(p); p = getchar())  sum = sum * 10 + p - 48;
    return sum * f;
}
const int maxn = 10001;
int v[maxn];
int n, m;
int hasjie[maxn][101];
int dp[maxn];
int main() {
    //freopen("test.txt", "r", stdin);
    n = read(), m = read();
    int sum = 0;
    for (int i = 1; i <= n; i++) {
        v[i] = read();
        sum += v[i];
    }
    sort(v + 1, v + n + 1,greater<int>());
    for (int i = 1; i <= n; i++) {
        for (int j = m; j >= v[i]; j--) {
            if (dp[j] <= dp[j - v[i]]+v[i]) {
                hasjie[i][j] = 1;
                dp[j] = dp[j - v[i]]+v[i];
            }
        }
    }
    if (dp[m] != m) {
        printf("No Solution\n");
    }
    else {
        int idx = n;
        vector<int>res;
        while (m > 0) {
            if (hasjie[idx][m]) {
                res.push_back(v[idx]);
                m -= v[idx];
            }
            idx--;
        }
        for (int i = 0; i < res.size(); i++) {
            if (i != 0)printf(" ");
            printf("%d", res[i]);
        }
    }
    return 0;
}