hdu 4796 Winter's Coming 插头DP

目前OJ上跑的最快 187ms
要求建造城墙把W分在左边，L分在右边，并且上下联通

做法是从地图第一行上面连一个插头下来。DP的最终状态是最后一行连一个插头到下面。
至于W分在左边，L分在右边，做法是：
保证任意一个W左边的下插头为偶数个，L的下插头为奇数个。

具体看代码

#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<math.h>
#include<string>
#include<cstdlib>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
void fre(){freopen("t.txt","r",stdin);}
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
#define ls o<<1
#define rs o<<1|1
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x, y) memcpy(x, y, sizeof(x))
typedef long long LL;
typedef unsigned int UI;
const int Z = 1e9+7,inf = 1e9;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a+=b; if(a >= Z) a%=Z; }
int cas;

const int MAXD=15,HASH=30007;
const int STATE=1000010;
int maze[25][MAXD];
int code[MAXD],ch[MAXD];
int n,m,ex,ey,ans;
struct HASHMAP
{
    int head[HASH],next[STATE],size;
    LL state[STATE];
    LL f[STATE];
    void init()
    {
        size=0;
        memset(head,-1,sizeof(head));
    }
    void push(LL st,LL ans)
    {
        int i;
        int h=st%HASH;
        for(i=head[h];i!=-1;i=next[i])
            if(state[i]==st)
            {
                gmin(f[i],ans);
                return;
            }
        state[size]=st;
        f[size]=ans;
        next[size]=head[h];
        head[h]=size++;
    }
}hm[2];
void decode(int *code,int m,LL  st)//解码
{
    for(int i=m;i>=0;i--)
    {
        code[i]=st&7;
        st>>=3;
    }
}
LL encode(int *code,int m)//状态压缩
{
    int cnt=1;
    MS(ch,-1);
    ch[0]=0;
    LL st=0;
    for(int i=0;i<=m;i++)
    {
        if(ch[code[i]]==-1)ch[code[i]]=cnt++;
        st<<=3;
        st|=ch[code[i]];
    }
    return st;
}
void shift(int *code,int m)//换行
{
    for(int i=m;i>0;i--)code[i]=code[i-1];
    code[0]=0;
}
void dpblank(int i,int j,int cur)
{
    int left,up;
    for(int 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].f[k]+maze[i][j]);
            }
        }
        else if(left || up)
        {
            int t;
            if(left) t=left;
            else t=up;
            if(maze[i][j+1] >= 0)
            {
                code[j-1]=0;
                code[j]=t;
                hm[cur^1].push(encode(code,m),hm[cur].f[k] + maze[i][j]);
            }
            if(maze[i+1][j] >= 0)
            {
                code[j-1]=t;
                code[j]=0;
                if(j==m)shift(code,m);
                hm[cur^1].push(encode(code,m),hm[cur].f[k] + maze[i][j]);
            }
        }
        else
        {
            if(j == m) shift(code,m);
            hm[cur^1].push(encode(code,m),hm[cur].f[k]);
            if(maze[i][j+1] >= 0&&maze[i+1][j] >= 0)
            {
                code[j-1]=code[j]=13;
                hm[cur^1].push(encode(code,m),hm[cur].f[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]);
        if(maze[i][j] == -3)//W的判断
        {
            int c = 0;
            for(int p = 0; p < j-1; ++p)
                if(code[p]) c++;
            if(c&1) continue;
        }
        if(maze[i][j] == -2)//L的判断
        {
            int c = 0;
            for(int p = 0; p < j-1; ++p)
                if(code[p]) c++;
            if(c%2 == 0) continue;
        }
        code[j-1]=code[j]=0;
        if(j==m) shift(code,m);
        hm[cur^1].push(encode(code,m),hm[cur].f[k]);
    }
}
void solve()
{
    int cur=0;
    ans= inf;
    hm[cur].init();

    MS(code,0);
    for(int i = 1; i <= m; ++i)//状态转移起点，从上面加一个插头下来
    {
        code[i-1] = 0; code[i] = 13;
        hm[cur].push(encode(code,m),0);
    }

    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        {
            hm[cur^1].init();
            if(maze[i][j] >= 0) dpblank(i,j,cur);
            else  dpblock(i,j,cur);
            cur^=1;
        }

    for(int i=0;i<hm[cur].size;i++)//统计答案，要求只有一个通往地图下面的插头
    {
        decode(code,m,hm[cur].state[i]);
        int c = 0;for(int i = 0; i <= m; ++i) if(code[i]) c++;
        if(c != 1) continue;
        gmin(ans,hm[cur].f[i]);
    }
    if(ans != inf) printf("%d\n",ans);
    else puts("-1");
}
char str[MAXD];
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        MS(maze,-1); ex=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%s",str+1);
            for(int j=1;j<=m;j++)
            {
                if(str[j] == '#') maze[i][j] = -1;
                else if(str[j] == 'L') maze[i][j] = -2;
                else if(str[j] == 'W') maze[i][j] = -3;
                else maze[i][j] = str[j] - '0';
            }
        }
        for(int i = 1; i <= m; ++i) maze[n+1][i] = 0;
        solve();
    }
    return 0;
}