hdoj 5617 Jam's maze（滚动数组dp求回文数）
Jam's maze
Problem Description
Jam got into a maze, he has to find a palindrome path from
(1,1) to
(N,N) to get out.
However he is not only thinking about how to get out of the maze,
but also counting the number of ways he can get out.
The maze a N∗N grid. On each cell is a uppercase letter.
Jam is at upper-left corner (1,1) now, and the exit is at lower-right corner (N,N).
Jam can only go down or go right.
Considering the number of palindrome path may be huge, you just need to output the number mod 5201314.
However he is not only thinking about how to get out of the maze,
but also counting the number of ways he can get out.
The maze a N∗N grid. On each cell is a uppercase letter.
Jam is at upper-left corner (1,1) now, and the exit is at lower-right corner (N,N).
Jam can only go down or go right.
Considering the number of palindrome path may be huge, you just need to output the number mod 5201314.
Input
The first line is
T(1≤T≤10) means
T Case.
For each case:
The first line is N(1≤N≤500) means the size of the maze.
Then N lines follow, each line has N uppercase letter.
For each case:
The first line is N(1≤N≤500) means the size of the maze.
Then N lines follow, each line has N uppercase letter.
Output
For each testcase, output the number of palindrome path mod
5201314.
Sample Input
1 4 ABCD BEFE CDEB GCBA
Sample Output
12Hintthere are 1 solutions is"ABCDCBA" there are 1 solutions is"ABCGCBA" there are 4 solutions is"ABEDEBA" there are 6 solutions is"ABEFEBA"
Source
BestCoder Round #70
]即可，状态转移很显然，如果相等的话把四个方向都扫一下即可。因为还是会爆空间，所以我们把第一维改成滚动数组就解决问题
#include<cstdio>
#include<cstring>
using namespace std;
const int MOD=5201314;
const int maxn=500+5;
char s[maxn][maxn];
int dp[2][maxn][maxn];
void add(int &x,int y)
{
x+=y;
if(x>=MOD)
x-=MOD;
}
int main()
{
int T,n,i,x1,x2,y1,y2;
scanf("%d",&T);
while(T--)
{
memset(dp,0,sizeof(dp));
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%s",s[i]+1);
if(s[n][n]!=s[1][1])
{
printf("0\n");
continue;
}
int step=n-1;
int op=0;
dp[0][1][n]=1;
for(i=1;i<=step;i++)
{
memset(dp[!op],0,sizeof(dp[!op]));
for(x1=1;x1<=i+1;x1++)
for(x2=n;x2>=n-i;x2--)
{
if(x1>x2)
continue;
y1=i+2-x1;
y2=n*2-x2-i;
if(s[x1][y1]==s[x2][y2])
{
add(dp[!op][x1][x2],dp[op][x1][x2+1]);
add(dp[!op][x1][x2],dp[op][x1][x2]);
add(dp[!op][x1][x2],dp[op][x1-1][x2]);
add(dp[!op][x1][x2],dp[op][x1-1][x2+1]);
}
}
op=!op;
}
int ans=0;
for(i=1;i<=n;i++)
add(ans,dp[op][i][i]);
printf("%d\n",ans);
}
return 0;
}