PAT 1045 Favorite Color Stripe

使用DP，递推关系见代码，先使用大空间的dp数组

#include <cstdio>

#include <cstdlib>

#include <vector>



using namespace std;



int max(int a, int b) {return a>b?a:b;}



int main() {

    int N, M, L, tmp;

    scanf("%d", &N);

    

    scanf("%d", &M);

    if (M < 1) return 0;

    vector<char> MColors(M);

    

    for (int i=0; i<M; i++) {

        scanf("%d", &tmp);

        MColors[i] = tmp;

    }

    

    scanf("%d", &L);

    if (L < 1) return 0;

    vector<char> LColors(L);

    

    for (int i=0; i<L; i++) {

        scanf("%d", &tmp);

        LColors[i] = tmp;

    }

    

    int rows = L + 1;

    int cols = M + 1;

    int total= rows * cols;

    int* dp = new int[rows * cols];

    

    for (int i=cols * (rows - 1); i<total; i++) {

        // set last row to zero;

        dp[i] = 0; 

    }

    for (int i=cols-1; i<total; i+=cols) {

        // set last column to zero;

        dp[i] = 0;

    }

    

    for (int i=rows-2; i>=0; i--) {

        int base = i * cols;

        for (int j=cols - 2; j>=0; j--) {

            int mc = MColors[j];

            int lc = LColors[i];

            if (mc == lc) {

                dp[base + j] = dp[base + j + cols] + 1;

            } else {

                dp[base + j] = max(dp[base + j + 1],

                                    dp[base + j + cols]);

            }

        }

    }

    printf("%d\n", dp[0]);

    return 0;

}

因为递推中只使用了当前行和下一行中的数据，题目又不要求给出完整解，因而可以优化空间如下：

#include <cstdio>

#include <cstdlib>

#include <vector>



using namespace std;



int max(int a, int b) {return a>b?a:b;}



int main() {

    int N, M, L, tmp;

    scanf("%d", &N);

    

    scanf("%d", &M);

    if (M < 1) return 0;

    vector<char> MColors(M);

    

    for (int i=0; i<M; i++) {

        scanf("%d", &tmp);

        MColors[i] = tmp;

    }

    

    scanf("%d", &L);

    if (L < 1) return 0;

    vector<char> LColors(L);

    

    for (int i=0; i<L; i++) {

        scanf("%d", &tmp);

        LColors[i] = tmp;

    }



    int cols = M + 1;

    vector<int> curr_row(cols, 0);

    vector<int> next_row(cols, 0);

    

    for (int i = L-1; i >=0; i--) {

        for (int j = M - 1; j>=0; j--) {

            if (MColors[j] == LColors[i]) {

                curr_row[j] = next_row[j] + 1;

            } else {

                curr_row[j] = max(next_row[j], curr_row[j + 1]);

            }

        }

        swap(curr_row, next_row);

    }

    printf("%d\n", next_row[0]);

    return 0;

}

即使用两个数组curr_row, next_row交替模拟整个dp数组