C语言动态规划(3)___Filthy Rich(Hdu 2391)

Problem Description

They say that in Phrygia, the streets are paved with gold. You’re currently on vacation in Phrygia, and to your astonishment you discover that this is to be taken literally: small heaps of gold are distributed throughout the city. On a certain day, the Phrygians even allow all the tourists to collect as much gold as they can in a limited rectangular area. As it happens, this day is tomorrow, and you decide to become filthy rich on this day. All the other tourists decided the same however, so it’s going to get crowded. Thus, you only have one chance to cross the field. What is the best way to do so?

Given a rectangular map and amounts of gold on every field, determine the maximum amount of gold you can collect when starting in the upper left corner of the map and moving to the adjacent field in the east, south, or south-east in each step, until you end up in the lower right corner.

 

Input

The input starts with a line containing a single integer, the number of test cases.
Each test case starts with a line, containing the two integers r and c, separated by a space (1 <= r, c <= 1000). This line is followed by r rows, each containing c many integers, separated by a space. These integers tell you how much gold is on each field. The amount of gold never negative.
The maximum amount of gold will always fit in an int.

 

Output

For each test case, write a line containing “Scenario #i:”, where i is the number of the test case, followed by a line containing the maximum amount of gold you can collect in this test case. Finish each test case with an empty line.

 

Sample Input

13 41 10 8 80 0 1 80 27 0 4

 

Sample Output

Scenario #1:42

 

题目大意：

输入第一行两个整数表示下面那个矩阵的行数和列数

第二行开始输入一个矩阵。

从左上角到右下角,只能向下或者向右走

输出所经过的数字和最大

这道题由于只能从两个方向我们可以使用DP，状态转移方程式就是

f [ i ] [ j ] = max{ f [ i-1 ] [ j ] , ,f [ i ] [ j-1 ] };

那么我们可以从第一排一直递推到最后一排，那么最后一个元素就是我们所求的最大。

代码如下：

#include <stdio.h>#include <string.h>int ans[1001][1001];     //dp数组int main(){    int N,i,j,m,n,ci=0,k;    scanf("%d",&N);    while(N--)    {        memset(ans,0,sizeof(ans));        ci++;        scanf("%d%d",&n,&m);        scanf("%d",&ans[0][0]);        for(i=1;i<m;i++)        {            scanf("%d",&k);            ans[0][i]=ans[0][i-1]+k;        }        for(i=1;i<n;i++)        {            scanf("%d",&k);               //每行第一列的元素只能由上一行第一列元素向下走，所以定值.            ans[i][0]=ans[i-1][0]+k;            for(j=1;j<m;j++)            {                scanf("%d",&k);                if(ans[i-1][j]>ans[i][j-1])         //状态转移过程                    ans[i][j]=ans[i-1][j]+k;                else                    ans[i][j]=ans[i][j-1]+k;            }        }        printf("Scenario #%d:\n",ci);        printf("%d\n",ans[n-1][m-1]);            printf("\n");    }    return 0;}