DP : Travel

最新推荐文章于 2019-05-12 18:35:11 发布

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本文探讨了一位旅行者如何在多个城市间旅行以最大化总收入的问题，通过每日选择前往另一城市或留在当前城市，同时考虑旅行费用和每日收入的变化。

Travel Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3230

Description

One traveler travels among cities. He has to pay for this while he can get some incomes.

Now there are n cities, and the traveler has m days for traveling. Everyday he may go to another city or stay there and pay some money. When he come to a city ,he can get some money. Even when he stays in the city, he can also get the next day's income. All the incomes may change everyday. The traveler always starts from city 1.

Now is your turn to find the best way for traveling to maximize the total income.

Input

There are multiple cases.

The first line of one case is two positive integers, n and m .n is the number of cities, and m is the number of traveling days. There follows n lines, one line n integers. The j integer in the i line is the expense of traveling from city i to city j. If i equals to j it means the expense of staying in the city.

After an empty line there are m lines, one line has n integers. The j integer in the i line means the income from city j in the i day.

The input is finished with two zeros.
n,m<100.

Output

You must print one line for each case. It is the max income.

Sample Input

Sample Output

Hint

In the Sample, the traveler can first go to city 2, then city 1, and finish his travel in city 1. The total income is:
-1+4-2+4-1+4=8;

#include <cstdio>
#include <cstring>
#include <cstdio>
#include <iostream>

using namespace std;

int c[110][110], v[110][110];
int f[110][110]; //第i天在j城时的最大收入

int main(){
    int n, m;
    int i, j, k;
    while(cin >> n >> m){
        if(n == 0 && m == 0)
        break;
        for(i = 1; i <= n; i ++)
        for(j = 1; j <= n; j ++){
            scanf("%d", &c[i][j]);  //i城到j城花费
        }
        for(i = 1; i <= m; i ++)
        for(j = 1; j <= n; j ++){
            scanf("%d", &v[i][j]);  //j城第i天收入
        }
        memset(f, -10, sizeof(f));
        f[0][1] = 0;
        for(i = 1; i <= m; i ++){
            for(j = 1; j <= n; j ++){
                for(k = 1; k <= n; k ++){
                    f[i][j] = max(f[i][j], f[i-1][k] + v[i][j] - c[k][j]);
                }
            }
        }
        int max = -9999999;
        for(i = 1; i <= n; i ++){
            if(max < f[m][i])
                max = f[m][i];
        }
        cout << max << endl;
    }
    return 0;
}