hdu3377之简单路径求最值

Plan

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 686    Accepted Submission(s): 234

Problem Description

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?

 

Input

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

 

Output

Output Format is "Case ID: ANS" one line for each data
Don't print any empty line to the output

 

Sample Input

2 21 23 13 30 -20 1001 -20 -201   1   1

 

Sample Output

Case 1: 5Case 2: 61

分析：

首先初始化状态即[1,0]的状态为1,表示从[1,0] => [1,1]有一个插头，这样就能保证是从点[1,1]开始然后将点[n+1,m]表示为可经过，然后进行插头DP时从[n,m] => [n+1,m]可以有插头，所以最终也保证了以点[n,m]结束在这个过程中进行插头DP只需要去掉p=1,q=2的情况就行，因为这样会形成环另外还有一种方法是增加外围使得题目变成求回路的最值0 0 0 0 00 # # # 0------1 0 1 |# 01 1 1 |# 0------# # 0 0 0这样增加外围就一定能保证求的回路是如果去掉外围就是从点[1,1]到[n,m]

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <queue>#include <algorithm>#include <map>#include <cmath>#include <iomanip>#define INF 99999999typedef long long LL;using namespace std;const int MAX=30000+10;const int N=10+10;int n,m,size,index;int val[N][N],total[2],bit[N];int head[MAX],next[MAX],Hash[MAX];LL dp[2][MAX],state[2][MAX],sum;void HashCalState(LL s,LL Val){    int pos=s%MAX;    for(int i=head[pos];i != -1;i=next[i]){        if(state[index][Hash[i]] == s){            dp[index][Hash[i]]=max(dp[index][Hash[i]],Val);            return;        }    }    ++total[index];    state[index][total[index]]=s;    dp[index][total[index]]=Val;    //头插法     Hash[size]=total[index];    next[size]=head[pos];    head[pos]=size++; }void DP(){    index=0;    total[index]=1;    state[index][1]=1;//初始化的状态是第一格有一个插头进来(从0进来)     dp[index][1]=val[1][1];    sum=-INF;    for(int i=1;i<=n;++i){        for(int j=1;j<=m;++j){            memset(head,-1,sizeof head);            size=0;            index=index^1;            total[index]=0;            for(int k=1;k<=total[index^1];++k){                LL s=state[index^1][k];                LL Val=dp[index^1][k];                int p=(s>>bit[j-1])%4;                int q=(s>>bit[j])%4;                int d=(s>>bit[j+1])%4;                if(!p && !q){                    HashCalState(s,Val);//跳过该格不经过(这里可以不用加判断if(i != n || j != m),因为n-1,m一定会经过n,m)                     if(val[i][j+1] == INF || val[i+1][j] == INF)continue;                    s=s+(1<<bit[j-1])+2*(1<<bit[j]);                    if(!d)HashCalState(s,Val+val[i+1][j]+val[i][j+1]+val[i][j]);//注意这里一定还要加上val[i][j]  else HashCalState(s,Val+val[i+1][j]+val[i][j]);//如果i,j+1已经由上面一行到达则不能重复累加该点价值                 }else if(!p && q){                    if(val[i][j+1] != INF){                    //if(!d)Val=Val+val[i][j+1];//这里不能这样改变Val,因为会影响下一个if里面的Val                    if(!d)HashCalState(s,Val+val[i][j+1]);                     else HashCalState(s,Val);//如果i,j+1已经由上面一行到达则不能重复累加该点价值                    }                    if(val[i+1][j] != INF){                        s=s+q*(1<<bit[j-1])-q*(1<<bit[j]);                        Val=Val+val[i+1][j];                        HashCalState(s,Val);                    }                }else if(p && !q){                    if(val[i+1][j] != INF)HashCalState(s,Val+val[i+1][j]);                    if(val[i][j+1] != INF){                        s=s-p*(1<<bit[j-1])+p*(1<<bit[j]);                        if(!d)HashCalState(s,Val+val[i][j+1]);                        else HashCalState(s,Val);//如果i,j+1已经由上面一行到达则不能重复累加该点价值                    }                }else if(p == 1 && q == 1){                    int b=1;                    for(int t=j+1;t<=m;++t){                        int v=(s>>bit[t])%4;                        if(v == 1)++b;                        if(v == 2)--b;                        if(!b){                            s=s+(1<<bit[t])-2*(1<<bit[t]);                            break;                        }                    }                    s=s-(1<<bit[j-1])-(1<<bit[j]);                    HashCalState(s,Val);                }else if(p == 2 && q == 2){                    int b=1;                    for(int t=j-2;t>=0;--t){                        int v=(s>>bit[t])%4;                        if(v == 2)++b;                        if(v == 1)--b;                        if(!b){                            s=s-(1<<bit[t])+2*(1<<bit[t]);                            break;                        }                    }                    s=s-2*(1<<bit[j-1])-2*(1<<bit[j]);                    HashCalState(s,Val);                }else if(p == 2 && q == 1){                    s=s-2*(1<<bit[j-1])-(1<<bit[j]);                    HashCalState(s,Val);                }            }        }        for(int k=1;k<=total[index];++k)state[index][k]<<=2;    }    sum=dp[index][1];//for(int k=1;k<=total[index];++k)sum=max(sum,dp[index][k]);//这里可以不用for，因为最后到达n,m->n+1,m状态只有一种 }int main(){    int num=0;    for(int i=0;i<N;++i)bit[i]=i<<1;    while(~scanf("%d%d",&n,&m)){        for(int i=1;i<N;++i)for(int j=1;j<N;++j)val[i][j]=INF;//这里i<=N,j<=N会覆盖bit的内存地址,教训啊         for(int i=1;i<=n;++i){            for(int j=1;j<=m;++j)scanf("%d",&val[i][j]);        }        val[n+1][m]=0;//在最后一个下面添加一个虚拟格子可以经过，则插头dp后的路线一定是以最后一格结尾         DP();        printf("Case %d: %lld\n",++num,sum);    }    return 0;}/*7 81  1  1  1  1  1  1  1-1 -1 -1 -1 -1 -1 -1  1 1  1  1 -1 -1 -1 -1  1 1 -1  1 -1 -1 -1 -1  1 1 -1  1  1  1  1  1  1 1 -1 -1 -1 -1 -1 -1 -1 1  1  1  1  1  1  1  1*/