标签：close scanf i++ stat open cas set mem ted
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 }
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 }
标签：close scanf i++ stat open cas set mem ted
原文地址：http://www.cnblogs.com/Aguin/p/6220802.html
