HDU 3377 Plan （插头DP)

Plan

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 449 Accepted Submission(s): 151


[align=left]Problem Description[/align]
One day, Resty comes to an incredible world to seek Eve -- The origin of life. Lilith, the sister of Eve, comes with him. Although Resty wants to find Eve as soon as possible, Lilith likes to play games so much that you can't make her make any move if you don't play with her.

Now they comes to the magical world and Lilish ask Resty to play with her.

The game is following :
Now the world is divided into a m * n grids by Lilith, and Lilith gives each grid a score.
So we can use a matrix to describe it.
You should come from cell(0, 0) to cell(m-1, n-1) (Up-Left to Down-Right) and try to colloct as more score as possible.
According to Lilish's rule, you can't arrive at each cell more than once.

Resty knows that Lilish will be easy to find the max score, and he doesn't want to lose the game.
So he want to find the game plan to reach the max score.

Your task is to calculate the max score that Lilish will find, the map is so small so it shouldn't be difficult for you, right?

[align=left]Input[/align]
The input consists of more than one testdata.
Process to the END OF DATA.
For each test data :
the first live give m and n. (1<=m<=8, 1<=n<=9)
following m lines, each contain n number to give you the m*n matrix.
each number in the matrix is between -2000 and 2000

[align=left]Output[/align]
Output Format is "Case ID: ANS" one line for each data
Don't print any empty line to the output

[align=left]Sample Input[/align]

2 2 1 2 3 1 3 3 0 -20 100 1 -20 -20 1 1 1

[align=left]Sample Output[/align]

Case 1: 5 Case 2: 61

[align=left]Author[/align]
Resty

[align=left]Source[/align]
HDOJ Monthly Contest – 2010.04.04

[align=left]Recommend[/align]
lcy

插头DP

/*
HDU 3377
从左上角走到右下角。每个格子有个分数。
每个格子只能经过一次，可以不经过
求最大分数
G++ 140ms
*/
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;

const int MAXD=15;
const int HASH=10007;
const int STATE=1000010;

int N,M;
int maze[MAXD][MAXD];
int score[MAXD][MAXD];
int code[MAXD];
int ch[MAXD];

struct HASHMAP
{
int head[HASH],state[STATE],next[STATE],size;
int dp[STATE];
void init()
{
size=0;
memset(head,-1,sizeof(head));
}
void push(int st,int ans)
{
int i,h=st%HASH;
for(i=head[h];i!=-1;i=next[i])
if(state[i]==st)
{
if(dp[i]<ans)dp[i]=ans;
return;
}
state[size]=st;
dp[size]=ans;
next[size]=head[h];
head[h]=size++;
}
}hm[2];
void decode(int *code,int m,int st)
{
for(int i=m;i>=0;i--)
{
code[i]=st&7;
st>>=3;
}
}
int encode(int *code,int m)
{
int cnt=1;
memset(ch,-1,sizeof(ch));
ch[0]=0;
int st=0;
for(int i=0;i<=m;i++)
{
if(ch[code[i]]==-1)ch[code[i]]=cnt++;
code[i]=ch[code[i]];
st<<=3;
st|=code[i];
}
return st;
}
void shift(int *code,int m)
{
for(int i=m;i>0;i--)code[i]=code[i-1];
code[0]=0;
}

void dpblank(int i,int j,int cur)
{
int k,left,up;
for(k=0;k<hm[cur].size;k++)
{
decode(code,M,hm[cur].state[k]);
left=code[j-1];
up=code[j];
if((i==1&&j==1)||(i==N&&j==M))
{
if((left&&(!up))||((!left)&&up))
{
code[j-1]=code[j]=0;
if(j==M)shift(code,M);
hm[cur^1].push(encode(code,M),hm[cur].dp[k]+score[i][j]);
}
else if(left==0&&up==0)
{
if(maze[i][j+1])
{
code[j-1]=0;
code[j]=13;
hm[cur^1].push(encode(code,M),hm[cur].dp[k]+score[i][j]);
}
if(maze[i][j+1])
{
code[j-1]=13;
code[j]=0;
if(j==M)shift(code,M);
hm[cur^1].push(encode(code,M),hm[cur].dp[k]+score[i][j]);
}
}
continue;
}
if(left&&up)
{
if(left==up)//没有这种情况，因为不形成环
{

}
else
{
code[j-1]=code[j]=0;
for(int t=0;t<=M;t++)//这里少了个等号，查了好久的错
if(code[t]==up)
code[t]=left;
if(j==M)shift(code,M);
hm[cur^1].push(encode(code,M),hm[cur].dp[k]+score[i][j]);
}
}
else if((left&&(!up))||((!left)&&up))
{
int t;
if(left)t=left;
else t=up;
if(maze[i][j+1])
{
code[j-1]=0;
code[j]=t;
hm[cur^1].push(encode(code,M),hm[cur].dp[k]+score[i][j]);
}
if(maze[i+1][j])
{
code[j]=0;
code[j-1]=t;
if(j==M)shift(code,M);
hm[cur^1].push(encode(code,M),hm[cur].dp[k]+score[i][j]);
}
}
else
{
if(maze[i][j+1]&&maze[i+1][j])
{
code[j]=code[j-1]=13;
hm[cur^1].push(encode(code,M),hm[cur].dp[k]+score[i][j]);
}
code[j-1]=code[j]=0;
if(j==M)shift(code,M);
hm[cur^1].push(encode(code,M),hm[cur].dp[k]);
}
}
}

void init()
{
memset(maze,0,sizeof(maze));
for(int i=1;i<=N;i++)
for(int j=1;j<=M;j++)
{
maze[i][j]=1;
scanf("%d",&score[i][j]);
}
}
void solve()
{
int i,j,cur=0;
hm[cur].init();
hm[cur].push(0,0);
for(int i=1;i<=N;i++)
for(int j=1;j<=M;j++)
{
hm[cur^1].init();
dpblank(i,j,cur);
cur^=1;
}
int ans=0;
for(int i=0;i<hm[cur].size;i++)
ans+=hm[cur].dp[i];
printf("%d\n",ans);
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int iCase=0;
while(scanf("%d%d",&N,&M)!=EOF)
{
iCase++;
printf("Case %d: ",iCase);
init();
if(N==1&&M==1)
{
printf("%d\n",score[1][1]);
continue;
}
solve();
}
return 0;
}

/*
Sample Input
2 2
1 2
3 1
3 3
0 -20 100
1 -20 -20
1   1   1

Sample Output
Case 1: 5
Case 2: 61

*/