ZOJ 3213 Beautiful Meadow

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简单路径，求经过的得到最大分数。
SOL：
老方法。
需要增加标志位记录独立插头个数，要使独立插头个数小于等于2.
#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;const int MAXD=15;const int HASH=419;const int STATE=1000010;int N,M;int maze[MAXD][MAXD];int code[MAXD];int ch[MAXD];int num;//独立插头的个数int ans;//答案struct HASHMAP{    int head[HASH],next[STATE],size;    int state[STATE],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){    num=st&7;//独立插头个数    st>>=3;    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];    }    st<<=3;    st|=num;    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(left&&up)        {            if(left!=up)            {                code[j-1]=code[j]=0;                for(int t=0;t<=M;t++)                  if(code[t]==up)                     code[t]=left;                if(j==M)shift(code,M);                hm[cur^1].push(encode(code,M),hm[cur].dp[k]+maze[i][j]);               // hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].dp[k]+maze[i][j]);            }        }        else if(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]+maze[i][j]);            }            if(maze[i+1][j])            {                code[j-1]=t;                code[j]=0;               hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].dp[k]+maze[i][j]);            }            if(num<2)            {                num++;                code[j-1]=code[j]=0;                hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].dp[k]+maze[i][j]);            }        }        else        {            code[j-1]=code[j]=0;           hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].dp[k]);            if(maze[i][j+1]&&maze[i+1][j])            {                code[j-1]=code[j]=13;                hm[cur^1].push(encode(code,M),hm[cur].dp[k]+maze[i][j]);            }            if(num<2)            {                num++;                if(maze[i][j+1])                {                    code[j]=13;                    code[j-1]=0;                    hm[cur^1].push(encode(code,M),hm[cur].dp[k]+maze[i][j]);                }                if(maze[i+1][j])                {                    code[j-1]=13;                    code[j]=0;                   hm[cur^1].push(encode(code,j==M?M-1:M),hm[cur].dp[k]+maze[i][j]);                }            }        }    }}void dpblock(int i,int j,int cur){    int k;    for(k=0;k<hm[cur].size;k++)    {        decode(code,M,hm[cur].state[k]);//这个忘记了！！！        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(){    scanf("%d%d",&N,&M);    ans=0;    memset(maze,0,sizeof(maze));//初始化别忘记了    for(int i=1;i<=N;i++)      for(int j=1;j<=M;j++)      {          scanf("%d",&maze[i][j]);          if(maze[i][j]>ans)ans=maze[i][j];      }}void solve(){    int i,j,cur=0;    hm[cur].init();    hm[cur].push(0,0);    for(i=1;i<=N;i++)       for(int j=1;j<=M;j++)       {           hm[cur^1].init();           if(maze[i][j])dpblank(i,j,cur);           else dpblock(i,j,cur);           cur^=1;       }    for(i=0;i<hm[cur].size;i++)    if(hm[cur].dp[i]>ans)        ans=hm[cur].dp[i];    printf("%d\n",ans);}int main(){    /*freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);*/    int T;    scanf("%d",&T);    while(T--)    {        init();        solve();    }    return 0;}
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