soj 1162: I-Keyboard(dp + 记忆化搜索)

@(K ACMer)

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题意:
手机有k个按键,你有l个字符需要印在这些按键上,每个字符被使用的平均频率已知.要求设计出一个最佳的按键分块方案,使期望的按次数最小.
分析:
就是把l个字符划分为k分,让每一份的按键次数加起来最小,典型的划分dp.
定义$dp[i][j]$为把前i个数划分为j个部分的最小期望按的次数总和.那么有转移方程:

$$dp[i][j] = min(dp[k][j - 1] + cal(k + 1, i))\ \ (j - 1 <= k < i)\ \ (cal(i, j)为直接计算i到j的期望)$$
这里cal(i , j)显然会被重复计算,故用记忆化搜索.

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Code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <set>
#include <map>
#include <stack>
#include <vector>
#include <string>
#include <queue>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
const int mod = int(1e9) + 7, INF = 0x3fffffff, maxn = 66 + 40;
pair<int, int> dp[maxn][maxn];
int k, l;
char key[maxn], a[maxn];
int p[maxn], dpcal[maxn][maxn];

int cal(int i, int j) {
    if (dpcal[i][j] != -1) return dpcal[i][j];
    int ret = 0;
    for (int k = i, l = 1; k <= j; k++, l++)
        ret += l * p[k];
    return dpcal[i][j] = ret;
}

int main(void) {
    int T;
    cin >> T;
    for (int cas = 1; cas <= T; cas++) {
        memset(dpcal, -1, sizeof(dpcal));
        cin >> k >> l;
        for (int i = 1; i <= k; i++)
            cin >> key[i];
        for (int j = 1; j <= l; j++)
            cin >> a[j];
        for (int i = 1; i <= l; i++)
            cin >> p[i];
        for (int i = 0; i <= l; i++)
            for (int j = 0; j <= k; j++)
                dp[i][j].first = INF;
        dp[0][0].first = 0;
        for (int i = 1; i <= l; i++) {
            for (int j = 1; j <= min(k, i); j++) {
                for (int m = i - 1; m >= j - 1; m--) {
                    if (j - 1 == 0 && m != 0) continue;
                    int temp = cal(m + 1, i);
                    if (dp[i][j].first >= dp[m][j - 1].first + temp) {
                        dp[i][j].first = dp[m][j - 1].first + temp;
                        dp[i][j].second = m;
                    }
                }
            }
        }
        stack<int> st;
        int t = dp[l][k].second, kk = k;
        st.push(l);
        while (st.size() < k) {
            st.push(t);
            t = dp[st.top()][--kk].second;
        }
        t = 1, kk = 1;

        cout << "Keypad #" << cas << ":" << endl;
        while (!st.empty()) {
            cout << key[t++] << ": ";
            for (; kk <= st.top(); kk++) {
                cout << a[kk];
            }
            cout << endl;
            st.pop();
        }
        cout << endl;
    }
    return 0;
}