HDU 1693 Eat the Trees 插头DP

　　这是一道入门题，只需判断插头有无。

　　具体分为：

　　　　1、上插头和左插头都有

　　　　2、有上插头或有左插头

　　　　3、上插头和左插头都没有

　　用HASHMAP储存状态，具体有一些小技巧（见程序）。 

  1 #include <cstdio>
  2 #include <cstdlib>
  3 #include <cstring>
  4 #include <string>
  5 #include <algorithm>
  6 
  7 using namespace std;
  8 
  9 #define REP(i, a, b) for (int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
 10 #define DWN(i, a, b) for (int i = (a), i##_end_ = (b); i >= i##_end_; --i)
 11 #define mset(a, b) memset(a, b, sizeof(a))
 12 typedef long long LL;
 13 const int MAXD = 15, HASH = 10007, STATE = 100010;
 14 int n, m, maze[MAXD][MAXD];
 15 int code[MAXD];
 16 struct HASHMAP
 17 {
 18     int head[HASH], nxt[STATE], siz, state[STATE]; LL f[STATE];
 19     void clear() { siz = 0, mset(head, -1); }
 20     void push(int x, LL add)
 21     {
 22         int pos = x%HASH, i = head[pos];
 23         for (; i != -1; i = nxt[i])
 24             if (state[i] == x)
 25             { f[i] += add; return ; }
 26         state[siz] = x, f[siz] = add;
 27         head[pos] = siz, nxt[siz++] = head[pos];    
 28     }
 29 }hm[2];
 30 
 31 void in()
 32 {
 33     scanf("%d %d", &n, &m), mset(maze, 0);
 34     REP(i, 1, n)
 35         REP(j, 1, m) scanf("%d", &maze[i][j]);
 36 }
 37 
 38 void decode(int s)
 39 {
 40     REP(i, 0, m)
 41         if (s&(1<<i)) code[i] = 1;
 42         else code[i] = 0;
 43 }
 44 
 45 int encode()
 46 {
 47     int ret = 0;
 48     DWN(i, m, 0) ret = ret*2+code[i];
 49     return ret;
 50 }
 51 
 52 void shift(int j) 
 53 { 
 54     if (j != m) return ;
 55     DWN(i, m, 1) code[i] = code[i-1]; code[0] = 0;
 56 }
 57 
 58 void dp_blank(int i, int j, int cur)
 59 {
 60     REP(k, 0, hm[cur].siz-1)
 61     {
 62         decode(hm[cur].state[k]);
 63         int lef = code[j-1], up = code[j];
 64         if (lef && up)
 65         {
 66             code[j-1] = code[j] = 0, shift(j);
 67             hm[cur^1].push(encode(), hm[cur].f[k]);
 68         }
 69         else
 70         {
 71             if (lef || up)
 72             {
 73                 if (maze[i+1][j])
 74                 {
 75                     code[j-1] = 1, code[j] = 0, shift(j);
 76                     hm[cur^1].push(encode(), hm[cur].f[k]);
 77                 }
 78                 if (maze[i][j+1])
 79                 {
 80                     code[j-1] = 0, code[j] = 1;
 81                     hm[cur^1].push(encode(), hm[cur].f[k]);
 82                 }
 83             }
 84             else
 85                 if (maze[i+1][j] && maze[i][j+1])
 86                 {
 87                     code[j-1] = code[j] = 1, shift(j);
 88                     hm[cur^1].push(encode(), hm[cur].f[k]);
 89                 }
 90         }
 91     }
 92 }
 93 
 94 void dp_block(int i, int j, int cur)
 95 {
 96     REP(k, 0, hm[cur].siz-1)
 97     {
 98         decode(hm[cur].state[k]);
 99         code[j-1] = code[j] = 0, shift(j);
100         hm[cur^1].push(encode(), hm[cur].f[k]);
101     }
102 }
103 
104 void work()
105 {
106     int cur = 0; LL ans = 0;
107     hm[0].clear(), hm[0].push(0, 1);
108     REP(i, 1, n)
109         REP(j, 1, m)
110         {
111             hm[cur^1].clear();
112             if (maze[i][j]) dp_blank(i, j, cur);
113             else dp_block(i, j, cur);
114             cur ^= 1;
115         }
116     REP(i, 0, hm[cur].siz-1) ans += hm[cur].f[i];
117     printf("There are %I64d ways to eat the trees.\n", ans);
118 }
119 
120 int main()
121 {
122     int T;
123     scanf("%d", &T);
124     REP(i, 1, T)
125     {
126         printf("Case %d: ", i);
127         in(), work();
128     }
129     return 0;
130 }
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posted @ 2017-02-23 19:26  Splay  阅读(263)  评论(0)  编辑  收藏  举报