HDU 2845 Beans（dp），hdu2845beansdp

HDU 2845 Beans（dp），hdu2845beansdp

Problem DescriptionBean-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 ? 
InputThere 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<=200000. 
OutputFor each case, you just output the MAX qualities you can eat and then get. 
Sample Input

4 611 0 7 5 13 978 4 81 6 22 41 40 9 34 16 1011 22 0 33 39 6

 
Sample Output

242

 
Source2009 Multi-University Training Contest 4 - Host by HDU 

思路：注意狀態轉移方程，還有就是行的轉移方程和列的相似，還有注意的是數組開大點

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define N 200005int dpx[N],dpy[N];int main(){    int n,m,i,j,s;    while(~scanf("%d%d",&n,&m))    {        memset(dpx,0,sizeof(dpx));        memset(dpy,0,sizeof(dpy));        for(i=2;i<=n+1;i++)        {             //memset(dpy,0,sizeof(dpy));             //這裡為什麼可以注釋掉呢，因為會被覆蓋                          for(j=2;j<=m+1;j++)            {                scanf("%d",&s);                dpy[j]=max(dpy[j-1],dpy[j-2]+s);  //dpy表示這一行從左至右能取到的最大的數和            }            dpx[i]=max(dpx[i-1],dpx[i-2]+dpy[1+m]);  //dpx表示從以上行中能取到數的最大的和        }        printf("%d\n",dpx[n+1]);    }    return 0;}

HDU2845 不知道錯了哪了，測試資料http://acmhdueducn/showproblemphp?pid=2845

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define max(a,b) (a)>(b)?(a):(b)
int gn[200020],map[200020];
int dist[200010],m,n;
int dp(int *a,int num)
{
memset(dist,0,sizeof(dist));
dist[1]=a[1];
for(int i=2;i<=num;i++)
dist[i]=max(dist[i-2]+a[i],dist[i-1]);
return dist[num];
}
int main()
{
int i,j,k,l;
while(scanf("%d%d",&m,&n)!=EOF)
{
memset(map,0,sizeof(map));
memset(gn,0,sizeof(gn));
for(i=1;i<=m;i++)
{
for(j=1;j<=n;j++)
{
scanf("%d",&gn[j]);
}
map[i]=dp(gn,n);
}
l=dp(map,m);
printf("%d\n",l);
}
}
拿著個程式去對拍就行了。
不知道Windows下怎麼對拍？網上查呀，對拍是一個很好的查錯的方法。
 
HDU1171 出現Runtime Error (ACCESS_VIOLATION) 具體看問題補充

while(~scanf("%d",&n),n!=-1) 整個輸入結束時，n是0，而不是-1：

while(n--)此迴圈結束後n的值是0