Unidirectional TSP UVA - 116

題目傳送門

思路：一個點可以往下延伸的狀態只有三種，我們可以得出一個遞推公式dp[i][j] = min(dp[i - 1][j - 1], dp[i][j - 1], dp[i + 1][j - 1])，dp[i][j]表示在(i, j)這個點上的最小值，但是我們還要求字典序最小的路徑，所以我們要對着過程倒序進行，然後記錄下來路徑。

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

#define MAXN 110
#define MAXE 110
#define INF 0x3f3f3f3f
#define MOD 100003
#define LL long long
#define ULL unsigned long long
#define pi 3.14159

using namespace std;

int n, m;
int box[MAXN][MAXN];
int dp[MAXN][MAXN];
int path[MAXN][MAXN];
int road[MAXN];

int solve(int i, int j) {
    if (i < 1 || i > n) {
        return INF;
    }
    int &ans = dp[i][j];
    if (ans < INF) {
        return ans;
    }
    if (j == m) {
        return ans = box[i][j];
    }
    if (i == 1) {
        int num1 = solve(i, j + 1);
        int num2 = solve(i + 1, j + 1);
        int num3 = solve(n, j + 1);
        if (num1 <= num2 && num1 <= num3) {
            path[i][j] = i;
            return ans = num1 + box[i][j];
        } else if (num2 <= num1 && num2 <= num3) {
            path[i][j] = i + 1;
            return ans = num2 + box[i][j];
        } else {
            path[i][j] = n;
            return ans = num3 + box[i][j];
        }
    } else if (i == n) {
        int num1 = solve(1, j + 1);
        int num2 = solve(i - 1, j + 1);
        int num3 = solve(i, j + 1);
        if (num1 <= num2 && num1 <= num3) {
            path[i][j] = 1;
            return ans = num1 + box[i][j];
        } else if (num2 <= num1 && num2 <= num3) {
            path[i][j] = i - 1;
            return ans = num2 + box[i][j];
        } else {
            path[i][j] = i;
            return ans = num3 + box[i][j];
        }
    } else {
        int num1 = solve(i - 1, j + 1);
        int num2 = solve(i, j + 1);
        int num3 = solve(i + 1, j + 1);
        if (num1 <= num2 && num1 <= num3) {
            path[i][j] = i - 1;
            return ans = num1 + box[i][j];
        } else if (num2 <= num1 && num2 <= num3) {
            path[i][j] = i;
            return ans = num2 + box[i][j];
        } else {
            path[i][j] = i + 1;
            return ans = num3 + box[i][j];
        }
    }
}

int main() {
    std::ios::sync_with_stdio(false);
    while (cin >> n >> m) {
        memset(dp, 0x3f3f3f3f, sizeof(dp));
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                cin >> box[i][j];
            }
        }
        for (int i = 1; i <= n; ++i) {
            solve(i, 1);
        }
        int x = 1, y = 1;
        int min_sum = INF;
        for (int i = 1; i <= n; ++i) {
            if (min_sum > dp[i][1]) {
                min_sum = dp[i][1];
                x = i;
            }
        }
        int cnt = 0;
        cout << x;
        x = path[x][y];
        y += 1;
        while (y <= m) {
            cout << " " << x;
            x = path[x][y];
            y += 1;
        }
        for (int i = cnt - 1; i >= 0; --i) {
            if (i == 0) {
                cout << road[i] << endl;
            } else {
                cout << road[i] << " ";
            }
        }
        cout << endl << min_sum << endl;
    }
    return 0;
}


/*
 5 6
 3 4 1 2 8 6
 6 1 8 2 7 4
 5 9 3 9 9 5
 8 4 1 3 2 6
 3 7 2 8 6 4
 5 6
 3 4 1 2 8 6
 6 1 8 2 7 4
 5 9 3 9 9 5
 8 4 1 3 2 6
 3 7 2 1 2 3
 2 2
 9 10
 9 10
 */