咸鱼好久啦。快放假了调整下。
圣诞打到了小礼物自己动手丰衣足食（其实是没人给我送）
12.24
URAL 1519 Formula 1
插头dp。抄板子。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 using namespace std;
6 typedef long long LL;
7 int N, M, ex, ey;
8 int G[22][22];
9
10 const int HASH = 10007;
11 const int STATE = 1111111;
12 struct HashMap
13 {
14 int sz, h[HASH];
15 int nxt[STATE];
16 LL key[STATE], val[STATE];
17
18 void init()
19 {
20 sz = 0;
21 memset(h, 0, sizeof(h));
22 }
23
24 void push(LL Key, LL Val)
25 {
26 int t = Key % HASH;
27 for(int i = h[t]; i; i = nxt[i])
28 if(key[i] == Key) {val[i] += Val; return;}
29 nxt[++sz] = h[t];
30 key[sz] = Key, val[sz] = Val;
31 h[t] = sz;
32 }
33 } H[2];
34
35 void decode(int * code, LL st)
36 {
37 for(int i = M; i >= 0; i--)
38 {
39 code[i] = st & 7LL;
40 st >>= 3LL;
41 }
42 }
43
44 int ch[22];
45 LL encode(int * code)
46 {
47 LL st = 0;
48 memset(ch, -1, sizeof(ch));
49 int cnt = ch[0] = 0;
50 for(int i = 0; i <= M; i++)
51 {
52 if(ch[code[i]] == -1) ch[code[i]] = ++cnt;
53 code[i] = ch[code[i]];
54 st <<= 3LL;
55 st |= (LL) code[i];
56 }
57 return st;
58 }
59
60 void shift(int * code)
61 {
62 for(int i = M; i; i--) code[i] = code[i-1];
63 code[0] = 0;
64 }
65
66 void solve(int x, int y, int cur)
67 {
68 H[cur^1].init();
69 for(int i = 1; i <= H[cur].sz; i++)
70 {
71 int code[22];
72 decode(code, H[cur].key[i]);
73 if(G[x][y])
74 {
75 int L = code[y-1], U = code[y];
76 if(L && U)
77 {
78 if(L == U)
79 {
80 if(x != ex || y != ey) continue;
81 code[y-1] = code[y] = 0;
82 }
83 else
84 {
85 code[y-1] = code[y] = 0;
86 for(int j = 0; j <= M; j++)
87 if(code[j] == U) code[j] = L;
88 }
89 }
90 else if(L || U)
91 {
92 int t = L ? L : U;
93 if(G[x][y+1])
94 {
95 code[y-1] = 0, code[y] = t;
96 H[cur^1].push(encode(code), H[cur].val[i]);
97 }
98 if(G[x+1][y])
99 {
100 code[y-1] = t, code[y] = 0;
101 if(y == M) shift(code);
102 H[cur^1].push(encode(code), H[cur].val[i]);
103 }
104 continue;
105 }
106 else
107 {
108 if(!G[x][y+1] || !G[x+1][y]) continue;
109 code[y-1] = code[y] = 20;
110 }
111 }
112 else code[y-1] = code[y] = 0;
113 if(y == M) shift(code);
114 H[cur^1].push(encode(code), H[cur].val[i]);
115 }
116 return;
117 }
118
119 int main(void)
120 {
121 ex = ey = 0;
122 scanf("%d %d", &N, &M);
123 for(int i = 1; i <= N; i++)
124 {
125 char s[22];
126 scanf("%s", s + 1);
127 for(int j = 1; j <= M; j++)
128 if(s[j] == ‘.‘) G[i][j] = 1, ex = i, ey = j;
129 }
130 if(!ex && !ey) puts("0");
131 else
132 {
133
134 int cur = 0;
135 H[cur].init();
136 H[cur].push(0, 1);
137 for(int i = 1; i <= N; i++)
138 for(int j = 1; j <= M; j++)
139 solve(i, j, cur), cur ^= 1;
140 LL ans = 0;
141 for(int i = 1; i <= H[cur].sz; i++) ans += H[cur].val[i];
142 printf("%I64d\n", ans);
143 }
144 return 0;
145 }
Aguin
12.25
HDU 1693 Eat the Trees
简化版。改改板子。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 using namespace std;
6 typedef long long LL;
7 int G[22][22];
8 int N, M;
9
10 const int HASH = 10007;
11 const int STATE = 1111111;
12 struct HashMap
13 {
14 int sz, h[HASH];
15 int nxt[STATE];
16 LL key[STATE], val[STATE];
17
18 void init()
19 {
20 sz = 0;
21 memset(h, 0, sizeof(h));
22 }
23
24 void push(LL Key, LL Val)
25 {
26 int t = Key % HASH;
27 for(int i = h[t]; i; i = nxt[i])
28 if(key[i] == Key) {val[i] += Val; return;}
29 nxt[++sz] = h[t];
30 key[sz] = Key, val[sz] = Val;
31 h[t] = sz;
32 }
33 } H[2];
34
35 void decode(int * code, LL st)
36 {
37 for(int i = M; i >= 0; i--)
38 {
39 code[i] = st & 7LL;
40 st >>= 3LL;
41 }
42 }
43
44 int ch[22];
45 LL encode(int * code)
46 {
47 LL st = 0;
48 memset(ch, -1, sizeof(ch));
49 int cnt = ch[0] = 0;
50 for(int i = 0; i <= M; i++)
51 {
52 if(ch[code[i]] == -1) ch[code[i]] = ++cnt;
53 code[i] = ch[code[i]];
54 st <<= 3LL;
55 st |= (LL) code[i];
56 }
57 return st;
58 }
59
60 void shift(int * code)
61 {
62 for(int i = M; i; i--) code[i] = code[i-1];
63 code[0] = 0;
64 }
65
66 void solve(int x, int y, int cur)
67 {
68 H[cur^1].init();
69 for(int i = 1; i <= H[cur].sz; i++)
70 {
71 int code[22];
72 decode(code, H[cur].key[i]);
73 if(G[x][y])
74 {
75 int L = code[y-1], U = code[y];
76 if(L && U)
77 {
78 if(L == U)
79 {
80 // if(x != ex || y != ey) continue;
81 code[y-1] = code[y] = 0;
82 }
83 else
84 {
85 code[y-1] = code[y] = 0;
86 for(int j = 0; j <= M; j++)
87 if(code[j] == U) code[j] = L;
88 }
89 }
90 else if(L || U)
91 {
92 int t = L ? L : U;
93 if(G[x][y+1])
94 {
95 code[y-1] = 0, code[y] = t;
96 H[cur^1].push(encode(code), H[cur].val[i]);
97 }
98 if(G[x+1][y])
99 {
100 code[y-1] = t, code[y] = 0;
101 if(y == M) shift(code);
102 H[cur^1].push(encode(code), H[cur].val[i]);
103 }
104 continue;
105 }
106 else
107 {
108 if(!G[x][y+1] || !G[x+1][y]) continue;
109 code[y-1] = code[y] = 20;
110 }
111 }
112 else code[y-1] = code[y] = 0;
113 if(y == M) shift(code);
114 H[cur^1].push(encode(code), H[cur].val[i]);
115 }
116 return;
117 }
118
119 int main(void)
120 {
121 int T;
122 scanf("%d", &T);
123 for(int kase = 1; kase <= T; kase++)
124 {
125 memset(G, 0, sizeof(G));
126 scanf("%d %d", &N, &M);
127 for(int i = 1; i <= N; i++)
128 for(int j = 1; j <= M; j++)
129 scanf("%d", &G[i][j]);
130 int cur = 0;
131 H[cur].init();
132 H[cur].push(0, 1);
133 for(int i = 1; i <= N; i++)
134 for(int j = 1; j <= M; j++)
135 solve(i, j, cur), cur ^= 1;
136 LL ans = 0;
137 for(int i = 1; i <= H[cur].sz; i++) ans += H[cur].val[i];
138 printf("Case %d: There are %I64d ways to eat the trees.\n", kase, ans);
139 }
140 return 0;
141 }
Aguin
时间： 2024-10-27 11:23:58