Maximum Vacation Days-LintCode

LintCode wants to give one of its best employees the option to travel among N cities to collect algorithm problems. But all work and no play makes Jack a dull boy, you could take vacations in some particular cities and weeks. Your job is to schedule the traveling to maximize the number of vacation days you could take, but there are certain rules and restrictions you need to follow.

Rules and restrictions:
1. You can only travel among N cities, represented by indexes from 0 to N-1. Initially, you are in the city indexed 0 on Monday.
2. The cities are connected by flights. The flights are represented as a N*N matrix (not necessary symmetrical), called flights representing the airline status from the city i to the city j. If there is no flight from the city i to the city j, flights[i][j] = 0; Otherwise, flights[i][j] = 1. Also, flights[i][i] = 0 for all i.
3. You totally have K weeks (each week has 7 days) to travel. You can only take flights at most once per day and can only take flights on each week's Monday morning. Since flight time is so short, we don't consider the impact of flight time.
4. For each city, you can only have restricted vacation days in different weeks, given an N*K matrix called days representing this relationship. For the value of days[i][j], it represents the maximum days you could take vacation in the city i in the week j.
You're given the flights matrix and days matrix, and you need to output the maximum vacation days you could take during K weeks.

 注意事项
N and K are positive integers, which are in the range of [1, 100].
In the matrix flights, all the values are integers in the range of [0, 1].
In the matrix days, all the values are integers in the range [0, 7].
You could stay at a city beyond the number of vacation days, but you should work on the extra days, which won't be counted as vacation days.
If you fly from the city A to the city B and take the vacation on that day, the deduction towards vacation days will count towards the vacation days of city B in that week.
We don't consider the impact of flight hours towards the calculation of vacation days.

样例
Given flights = [[0,1,1],[1,0,1],[1,1,0]], days = [[1,3,1],[6,0,3],[3,3,3]], return 12.

Explanation: 
Ans = 6 + 3 + 3 = 12. 

One of the best strategies is:
1st week : fly from city 0 to city 1 on Monday, and play 6 days and work 1 day. 
(Although you start at city 0, we could also fly to and start at other cities since it is Monday.) 
2nd week : fly from city 1 to city 2 on Monday, and play 3 days and work 4 days.
3rd week : stay at city 2, and play 3 days and work 4 days.

Given flights = [[0,0,0],[0,0,0],[0,0,0]], days = [[1,1,1],[7,7,7],[7,7,7]], return 3.

Explanation: 
Ans = 1 + 1 + 1 = 3. 

Since there is no flights enable you to move to another city, you have to stay at city 0 for the whole 3 weeks. 
For each week, you only have one day to play and six days to work. 
So the maximum number of vacation days is 3.

Given flights = [[0,1,1],[1,0,1],[1,1,0]], days = [[7,0,0],[0,7,0],[0,0,7]], return 21.

Explanation:
Ans = 7 + 7 + 7 = 21

One of the best strategies is:
1st week : stay at city 0, and play 7 days. 
2nd week : fly from city 0 to city 1 on Monday, and play 7 days.
3rd week : fly from city 1 to city 2 on Monday, and play 7 days.

思路
题目的意思就是在给定了飞机航线以及每个城市每周休息时间的情况下，求最大休息时间的问题。

#ifndef C874_H
#define C874_H
#include<iostream>
#include<vector>
using namespace std;
class Solution {
public:
    /**
    * @param flights: the airline status from the city i to the city j
    * @param days: days[i][j] represents the maximum days you could take vacation in the city i in the week j
    * @return: the maximum vacation days you could take during K weeks
    */
    int maxVacationDays(vector<vector<int>> &flights, vector<vector<int>> &days) {
        // Write your code here
        int rows = days.size();
        int cols = days[0].size();
        //为了方便计算,将flights[i][i]置为1,表示可以到达
        for (int i = 0; i < flights.size(); ++i)
            flights[i][i] = 1;
        //dp[i][j]表示在第j周在城市i的情况下,累积最大的休息天数
        //初始值为-1,表示无法在第j周到达城市i
        vector<vector<int>> dp(rows, vector<int>(cols, -1));
        //由于开始位置0号城市,计算0周的最大休息天数
        for (int i = 0; i < rows; ++i)
        {
            if (flights[0][i] == 1)
            {
                dp[i][0] = days[i][0];
            }
        }
        //由于dp[i][j]表示在第j周在城市i的情况下,累积最大的休息天数
        //遍历j的前一列,求出从每个城市到i城市最大的休息天数
        //当飞机无法到达或者前一列元素值为-1时,表示此种情况无法满足,跳过
        for (int j = 1; j < cols; ++j)
        {
            for (int i = 0; i < rows; ++i)
            {
                int res = -1;
                for (int k = 0; k < rows; ++k)
                {
                    if (flights[k][i] == 1&&dp[k][j-1]!=-1)
                    {
                        res = maxVal(res, dp[k][j - 1] + days[i][j]);
                    }
                }
                dp[i][j] = res;
            }
        }
        int num = 0;
        for (int i = 0; i<rows; ++i)
        {
            if (dp[i][cols - 1]>num)
                num = dp[i][cols - 1];
        }
        return num;
    }
    int maxVal(int a, int b)
    {
        return a>b ? a : b;
    }
};
#endif