（难-需再看）1068 Find More Coins （30 分）

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 10​4​​ 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 (≤10​4​​, the total number of coins) and M (≤10​2​​, 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

题意：

 有N枚硬币，给出每枚硬币的价值，现在要用这些硬币去支付价值为M的东西，问是否可以找到这样的方案，使得选择用来支付的硬币价值之和恰好为M。如果不存在，输出No Solution；如果存在，从小到大输出选择支付的硬币价格，如果有多个方案，输出字典序最小的那个

思路：

背包问题

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
const int maxn = 10010;
const int maxv = 110;
int w[maxn], dp[maxv] = {0};    //w[i]存储钱币的价值
bool choice[maxn][maxv], flag[maxn];    //choice记录dp[i][v]时选择了哪个策略
bool cmp(int a, int b) {   //从大到小排序
    return a > b;
}
int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &w[i]);
    }
    sort(w + 1, w + n + 1, cmp);
    for(int i = 1; i <= n; i++) {
        for(int v = m; v >= w[i]; v--) {
            //状态转移方程
            if(dp[v] <= dp[v - w[i]] + w[i]) {      
                dp[v] = dp[v - w[i]] + w[i];    
                choice[i][v] = 1;       //放入第i件物品
            } else {
                choice[i][v] = 0;       //不放第i件物品
            }
        }
    }
    if(dp[m] != m) {                    //无解
        printf("No Solution");
    } else {
        //记录最优路径
        int k = n, num = 0, v = m;
        while(k >= 0) {
            if(choice[k][v] == 1) {
                flag[k] = true;
                v -= w[k];
                num++;
            } else {
                flag[k] = false;
            }
            k--;
        }
        //输出方案
        for(int i = n; i >= 1; i--) {
            if(flag[i] == true) {
                printf("%d", w[i]);
                num--;
                if(num > 0) {
                    printf(" ");
                }
            }
        }
    }
    return 0;
}