poj 3230 Travel（dp）-白红宇

poj 3230 Travel（dp）

阅读量：6002 次

发布时间：2019-06-20

本文共 2589 字，大约阅读时间需要 8 分钟。

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

3 33 1 22 3 11 3 22 4 34 3 23 4 20 0

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;

dp[i][j] 代表第i天在j城市最多赚了多少钱。

起点在1，所以dp[0][1]=0

然后三重循环dp就好了·

注意赚的钱有可能是负的~

1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include
       
         3 #include
        
          4 #include
         
           5 #include
          
            6 #include
           
             7 #include
            
              8 #include
             
               9 #include
              
               10 #include
               
                11 #include
                12 #include
                 
                  13 #include
                  
                   14 using namespace std;15 #define ll long long16 #define eps 1e-1017 #define MOD 100000000718 #define N 10619 #define inf 1<<2620 int n,m;21 int cost[N][N];22 int _get[N][N];23 int dp[N][N];24 int main()25 {26 while(scanf("%d%d",&n,&m)==2){27 if(n==0 && m==0){28 break;29 }30 for(int i=1;i<=n;i++){31 for(int j=1;j<=n;j++){32 scanf("%d",&cost[i][j]);33 }34 }35 for(int i=1;i<=m;i++){36 for(int j=1;j<=n;j++){37 scanf("%d",&_get[i][j]);38 }39 }40 for(int i=0;i<=m;i++){41 for(int j=0;j<=n;j++){42 dp[i][j]=-inf;43 }44 }45 dp[0][1]=0;//第0天，从城市1开始，值为046 for(int i=1;i<=m;i++){47 for(int j=1;j<=n;j++){48 for(int k=1;k<=n;k++){49 if(_get[j][k]<0)continue;50 dp[i][k]=max(dp[i][k],dp[i-1][j]-cost[j][k]+_get[i][k]);51 }52 }53 } 54 55 int maxxx=-inf;56 for(int i=1;i<=n;i++){57 maxxx=max(maxxx,dp[m][i]);58 }59 printf("%d\n",maxxx);60 }61 return 0;62 }