USACO 1.4 clocks 分析

Description：就不囉嗦了，見USACO的題目地址吧

http://ace.delos.com/usacoprob2?a=GYKcgnGbqEV&S=clocks

Analysis：

由於同一種方法轉4次相當於不轉，所以轉的方案數共有4^9種；

9個鐘的狀態也共有4^9種；

分別用1，2，3，0來表示3，6，9，12則最終狀態爲全0；

 

枚舉：分別枚舉9種方案的使用次數，然後判斷是否可行，下面的這個程序給我了很多啓發

/* ID: MasterRay LANG: C++ TASK: clocks Executing... Test 1: TEST OK [0.022 secs, 2884 KB] Test 2: TEST OK [0.022 secs, 2884 KB] Test 3: TEST OK [0.022 secs, 2884 KB] Test 4: TEST OK [0.022 secs, 2884 KB] Test 5: TEST OK [0.011 secs, 2884 KB] Test 6: TEST OK [0.011 secs, 2884 KB] Test 7: TEST OK [0.022 secs, 2884 KB] Test 8: TEST OK [0.022 secs, 2884 KB] Test 9: TEST OK [0.022 secs, 2884 KB] */ #include <algorithm> #include <climits> #include <cstdio> #include <numeric> using namespace std; #define LOOP(x) for (s[x] = 0; s[x] < 4; ++s[x]) //很少用啊，強大 int main() { int a[9], b[9], cnt, min_cnt = INT_MAX, res[10], s[10]; freopen("clocks.in", "r", stdin); freopen("clocks.out", "w", stdout); for (int i = 0; i < 9; ++i) scanf("%d", &a[i]), a[i] = a[i]/3%4;//使用了逗號表達式適當精簡代碼 LOOP(1) LOOP(2) LOOP(3) LOOP(4) LOOP(5) LOOP(6) LOOP(7) LOOP(8) LOOP(9) { b[0] = (a[0]+s[1]+s[2]+s[4])%4; b[1] = (a[1]+s[1]+s[2]+s[3]+s[5])%4; b[2] = (a[2]+s[2]+s[3]+s[6])%4; b[3] = (a[3]+s[1]+s[4]+s[5]+s[7])%4; b[4] = (a[4]+s[1]+s[3]+s[5]+s[7]+s[9])%4; b[5] = (a[5]+s[3]+s[5]+s[6]+s[9])%4; b[6] = (a[6]+s[4]+s[7]+s[8])%4; b[7] = (a[7]+s[5]+s[7]+s[8]+s[9])%4; b[8] = (a[8]+s[6]+s[8]+s[9])%4; if (accumulate(b, b+9, 0)) continue;//用了一個小函數 cnt = accumulate(s+1, s+10, 0); if (cnt < min_cnt) { min_cnt = cnt; copy(s+1, s+10, res+1); } } for (int i = 1; i < 10; ++i) while (res[i]--) printf("%d%c", i, --cnt ? ' ' : '/n');//用了3目運算符 }

 

 

Breadth First Search

這是我做的方法，一開始沒有用位運算來加速，所有就用了一個結構來保存，第一次用了結構內聲明函數來計算hash值；

/* ID: zhangji42 PROG: clocks LANG: C++ 不用非遞減拓展 Executing... Test 1: TEST OK [0.032 secs, 15432 KB] Test 2: TEST OK [0.032 secs, 15432 KB] Test 3: TEST OK [0.076 secs, 15432 KB] Test 4: TEST OK [0.086 secs, 15432 KB] Test 5: TEST OK [0.108 secs, 15432 KB] Test 6: TEST OK [0.302 secs, 15432 KB] Test 7: TEST OK [0.378 secs, 15432 KB] Test 8: TEST OK [0.378 secs, 15432 KB] Test 9: TEST OK [0.378 secs, 15432 KB] 使用非遞減拓展 Executing... Test 1: TEST OK [0.043 secs, 15432 KB] Test 2: TEST OK [0.043 secs, 15432 KB] Test 3: TEST OK [0.054 secs, 15432 KB] Test 4: TEST OK [0.032 secs, 15432 KB] Test 5: TEST OK [0.054 secs, 15432 KB] Test 6: TEST OK [0.097 secs, 15432 KB] Test 7: TEST OK [0.108 secs, 15432 KB] Test 8: TEST OK [0.097 secs, 15432 KB] Test 9: TEST OK [0.086 secs, 15432 KB] ABCDEFGHI 012345678 */ #include <stdio.h> #include <string.h> #include <ctype.h> #include <time.h> const int maxlen = 262144+3; struct INFO { short c[9], type; int hash, pre; int gethash(){ //內置函數，面向對象的手法，不過這裏用的挺好，不顯累贅 hash = 0; for (int i = 0; i < 9; i++) hash = hash*4+c[i]; return hash; } INFO(){ pre = 1 << 20; hash = maxlen-1; type = 1;} //函數初始化 }; INFO q[maxlen]; bool h[maxlen]; INFO move(INFO q, int type, int pre) { q.pre = pre; q.type = type; char ch[7] = ""; switch (type) { case 1:strcpy(ch, "ABDE");break; case 2:strcpy(ch,"ABC");break; case 3:strcpy(ch,"BCEF");break; case 4:strcpy(ch,"ADG");break; case 5:strcpy(ch,"BDEFH");break; case 6:strcpy(ch,"CFI");break; case 7:strcpy(ch,"DEGH");break; case 8:strcpy(ch,"GHI");break; case 9:strcpy(ch,"EFHI");break; } for (int i = 0; i < strlen(ch); i++) q.c[ch[i]-'A'] = (q.c[ch[i]-'A']+1)%4; q.gethash(); return q; } void print(int now) { if (q[now].pre == 0) { printf("%d", q[now].type); return; } print(q[now].pre); printf(" %d", q[now].type); } /*for test:print the clock void printtest(INFO q) { for (int i = 0; i < 9; i++) { if (q.c[i] == 0) printf("12 "); else printf("%d ", q.c[i]*3); if (i % 3 == 2) printf("/n"); } printf("/n"); } */ int main(){ freopen("clocks.in", "r", stdin); freopen("clocks.out", "w", stdout); for (int i = 0; i < 9; i++) scanf("%d", &q[0].c[i]),q[0].c[i] = q[0].c[i]/3%4; memset(h,0,sizeof(h)); h[q[0].gethash()] = 1; /*Breath First Search*/ int now = 0, full = 0; INFO newget; if (!h[0]) while (now <= full) { for (int i = q[now].type; i <= 9; i++) { //使用了非遞減拓展，大大優化了效率 newget = move(q[now], i, now); //printtest(newget);//to test whether 9 move is correct if (!h[newget.hash]) { q[++full] = newget; h[newget.hash] = 1; } if (newget.hash == 0) break; } //break;//test if (!newget.hash) break; now++; } /*print the path*/ if (full > 0) { print(full); printf("/n"); } //printf("Time used:%.4lf/n", (double)clock() / CLOCKS_PER_SEC); return 0; }   

位運算加速BFS

後來看了NOCOW上的位運算的思路，“

這樣，它們對應的二進制數爲：000，001，010，011。即，我們用三個位來記錄一個時鐘的狀態（爲什麼不用兩位？請思考）。

 

要tick一個時鐘的時候，就給該位加上一，再用按位與的方法去除高位的1。

 

令最高的三位爲時鐘A，最低的三位爲時鐘I，那麼：

 

數“57521883”用於清除每個時鐘狀態最高位的一（按位與）。

 

const long move[9] = {18911232, 19136512, 2363904, 16810048, 2134536, 262657, 36936, 73, 4617}

 

move[i]表示題述中的第i+1種方法

 

令f[q]爲原狀態，比如用題述中的第k種方法，那麼可以寫成 f[q + 1] = (f[q] + move[k - 1]) & 57521883;

 

當9個時鐘都回歸12點的時候，巧的是狀態f=0。這樣，判斷每個狀態f是否爲0，就知道是否求出可行解。

“

上述方法要想用C++的程序過是不太可行的（除非你用要麻煩一下，把一個數拆成9個去hash，但這樣就體現不出位運算的優勢了），因爲要開一個57521883大的BOOL數組，但在C++中空間分配至少是1個字節，BOOL的前7位是無用的0，所以大小達到了57521883/2^20 M = 54 M，如果能用1位來存貯，則使用54 / 8 M = 6.75 M，由於USACO評測系統給的內存只有16 M，所以就不行了，但用自己電腦測試還是可以的，速度比前一種方法快10%左右。

還要特別注意的是：<<和+的優先級，比如 1 << 2 + 3 = ？，不是7，而是32，因爲 << 的優先級比 +/- 低

/* ID: zhangji42 PROG: clocks LANG: C++ */ #include <stdio.h> #include <string.h> #include <time.h> const int move[9] = {18911232, 19136512, 2363904, 16810048, 2134536, 262657, 36936, 73, 4617}, maxlen = 262144+3; struct INFO { short type; int hash, pre; INFO(){ pre = 0; hash = 0; type = 0;} }; INFO q[maxlen]; bool h[57521883+1];//6.75M void print(int now) { if (q[now].pre == 0) { printf("%d", q[now].type); return; } print(q[now].pre); printf(" %d", q[now].type); } int main() { freopen("clocks.in", "r", stdin); freopen("clocks.out", "w", stdout); int clo; for (int i = 0; i < 9; i++) { scanf("%d", &clo); q[0].hash = q[0].hash << 3; q[0].hash += (clo / 3) % 4; } q[0].type = 1; memset(h,0,sizeof(h)); h[q[0].hash] = 1; int now = 0, full = 0, newhash; if (!h[0]) while (now <= full) { for (int i = q[now].type; i <= 9; i++) { newhash = (q[now].hash + move[i-1]) & 57521883; if (!h[newhash]) { q[++full].hash = newhash; q[full].pre = now; q[full].type = i; h[newhash] = 1; } if (!newhash) break; } if (!newhash) break; now++; } //printf("%d/n", now); if (full) { print(full); printf("/n"); } //printf("Time used:%.4lf/n", (double)clock() / CLOCKS_PER_SEC); return 0; }

還有一種很強的數學方法，和解方程法；