POJ 1742 Coins

题意：现有n种面额的硬币，面额、个数分别为A_i、C_i，要求最多能给出几种不超过m的金额。
分析：这是一个多重部分和问题(多重背包问题)，最基本的做法是：

dp[i][j] := 前i种硬币能否凑成j

递推关系式：

dp[i][j]=1 (存在k，使dp[i-1][j-k*a[i]]=1，其中k<=c[i]、k*a[i]<=j)

#include<iostream>

using namespace std;

int dp[105][100005];
int a[105];
int c[105];

int main() {
    int n, m;
    cin >> n >> m;
    memset(dp, 0, sizeof(dp));

    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }
    for (int i = 0; i < n; ++i) {
        cin >> c[i];
    }

    dp[0][0] = 1;
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j <= m; ++j) {
            for (int k = 0; k <= c[i] && k*a[i] <= j; ++k) {
                dp[i + 1][j] |= dp[i][j - k*a[i]];
            }
        }
    }
    int count = 0;
    for (int i = 1; i <= m; ++i) {
        if (dp[n][i]) {
            count++;
        }
    }
    cout << count << endl;

    return 0;
}

dp[i][j] := 用前i种硬币凑成j时，第i种硬币最多剩余数（-1表示配不出来）

1. 如果dp[i - 1][j] >= 0(前i-1个数可以凑出j，那么第i个数根本用不着)则dp[i][j]=C[i],
2. 如果j < a[i]或者dp[i][j - a[i]] <=0 (面额太大或者在配更小的数的时候就用光了）dp[i][j] = -1,
3. 其他（将第i个数用掉一个）dp[i][j] = dp[i][j-a[i]] - 1

#include<iostream>

using namespace std;

int dp[105][100005];
int a[105];
int c[105];

int main() {
    int n, m;
    while ((cin >> n >> m) && n + m) {
        memset(dp, -1, sizeof(dp));

        for (int i = 0; i < n; ++i) {
            cin >> a[i];
        }
        for (int i = 0; i < n; ++i) {
            cin >> c[i];
        }

        dp[0][0] = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j <= m; ++j) {
                if (dp[i][j] >= 0) {
                    dp[i + 1][j] = c[i];
                }
                else if (j < a[i] || dp[i + 1][j - a[i]] <= 0) {
                    dp[i + 1][j] = -1;
                }
                else {
                    dp[i + 1][j] = dp[i + 1][j - a[i]] - 1;
                }
            }
        }
        int count = 0;
        for (int i = 1; i <= m; ++i) {
            if (dp[n][i] != -1) {
                count++;
            }
        }
        cout << count << endl;
    }

    return 0;
}

利用一维dp数组：

#include<iostream>

using namespace std;

int dp[100005];
int a[105];
int c[105];

int main() {
    int n, m;
    while ((cin >> n >> m) && n + m) {
        memset(dp, -1, sizeof(dp));

        for (int i = 0; i < n; ++i) {
            cin >> a[i];
        }
        for (int i = 0; i < n; ++i) {
            cin >> c[i];
        }

        dp[0] = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j <= m; ++j) {
                if (dp[j] >= 0) {
                    dp[j] = c[i];
                }
                else if (j < a[i] || dp[j - a[i]] <= 0) {
                    dp[j] = -1;
                }
                else {
                    dp[j] = dp[j - a[i]] - 1;
                }
            }
        }
        int count = 0;
        for (int i = 1; i <= m; ++i) {
            if (dp[i] != -1) {
                count++;
            }
        }
        cout << count << endl;
    }

    return 0;
}

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参考：POJ 1742 Coins 图表详解 《挑战程序设计竞赛(第2版)》