nyoj-动态规划-234-吃土豆-201308131021

吃土豆

时间限制：1000 ms | 内存限制：65535 KB

难度：4

描述 Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled with different qualities beans. Meantime, there is only one bean in any 1*1 grid. Now you want
to eat the beans and collect the qualities, but everyone must obey by the following rules: if you eat the bean at the coordinate(x, y), you can’t eat the beans anyway at the coordinates listed (if exiting): (x, y-1), (x, y+1), and the both rows whose abscissas
are x-1 and x+1.



Now, how much qualities can you eat and then get ?

输入There are a few cases. In each case, there are two integer M (row number) and N (column number). The next M lines each contain N integers, representing the qualities of the beans. We can make sure that the quality of bean isn't beyond 1000, and 1<=M,N<=500.

输出For each case, you just output the MAX qualities you can eat and then get.样例输入

4 6
11 0 7 5 13 9
78 4 81 6 22 4
1 40 9 34 16 10
11 22 0 33 39 6

样例输出

242

先说一下此题的大意：在一个二维数组中保存各个土豆的重量，每次取一个土豆，它上一行与下一行均不能选择，还有它左右的两个也不能选，按照这个规则遍历整个二维数组，最后输出我们所能获得的最大土豆质量；

思路如下：

典型的递归分制与递推问题的处理，此题若用递归算法代码将更简洁，只可惜由于时间限制，它会出现超时现象，所以我们要把它转化成非递归的递推算法，但递归思想我们不得不去领悟的，故笔者将对应的递归算法代码也一起分享给大家，希望对大家有所帮助，废话不多说，我们来看这题的解决办法：

通过分析题目我们不难发现，我们可以先对每一行求对应的最大统计值，在求解过程中注意，每取一个值，与之左右相邻的两个值均不能选取，在我们求前i个数的最大统计值时，我们要先确定前i-1个数的最大统计值，仔细体会这个方程f[i]=f[i-2]+b[i]>f[i-1]?f[i-2]+b[i]:f[i-1];，下面给出具体代码：

#include<stdio.h>
/*递归分制算法，
int max(int n,int *b){
if(n==1)return b
;
if(n==2)return b
>b[n-1]?b
:b[n-1];
return max(n-2,b)+b
>max(n-1,b)?max(n-2,b)+b
:max(n-1,b);
}*/
////非递归递推算法///////////
int max(int n,int *b){
int i,f[510];
for(i=1;i<=n;i++)
{
if(i==1)f[i]=b[i];
else if(i==2)
f[i]=b[i]>b[i-1]?b[i]:b[i-1];
else
{
f[i]=f[i-2]+b[i]>f[i-1]?f[i-2]+b[i]:f[i-1];
}//else
}//for
return f
;
}
int main()
{
int a[510],b[510];
int i,j,M,N;
while(scanf("%d %d",&M,&N)!=EOF)
{
for(i=1;i<=M;i++)
{
for(j=1;j<=N;j++)
{
scanf("%d",&a[j]);
}
b[i]=max(N,a);
}//for
printf("%d\n",max(M,b));
}//while
//system("pause");
return 0;
}

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