矩陣從左上角向右下角走，每次只能向右或者向下移動，求經過最小的路徑

先通過動態規划求出最小路徑的值，然后根據dp二維數組倒推所走路徑。參考找出最大公共子序列解法。

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

int minsum(vector<vector<int>>a, vector<int>&reg){
    vector<vector<int>>dp(a.size(), vector<int>(a[0].size(), a[0][0]));
    
    for (int i = 1; i < a.size(); i++){
        dp[i][0] = dp[i - 1][0]+a[i][0];
    }
    for (int j = 1; j < a[0].size(); j++){
        dp[0][j] = dp[0][j - 1] + a[0][j];
    }
    for (int i = 1; i < a.size(); i++){
        for (int j = 1; j < a[0].size(); j++){
            dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + a[i][j];
        }
    }
    /*for (int i = 0; i < dp.size(); i++){
        for (int j = 0; j < dp[0].size(); j++){
            cout << dp[i][j] << " ";
        }
        cout << endl;
    }*/
    
    int i = dp.size() - 1;
    int j = dp[0].size() - 1;
    while (i >= 0 && j >= 0){
        reg.push_back(a[i][j]);
        if (i > 0 && j > 0){
            if (dp[i][j - 1] > dp[i - 1][j]){
                i--;
            }
            else{
                j--;
            }
            continue;
        }
        else if (i == 0){
            j--;
        }
        else{
            i--;
        }
    }
    return dp[a.size()-1][a[0].size()-1];
}

int main(){
    vector<vector<int>>a = { { 1, 3, 1 }, { 1, 5, 1 }, { 4, 2, 1 } };
    vector<int>reg;
    int res = minsum(a,reg);
    reverse(reg.begin(), reg.end());
    for (int i = 0; i < reg.size(); i++){
        cout << reg[i] << endl;
    }

    system("pause");
    return 0;
}