【USACO3.1.3】丑数 恶心搜索题

TLE方法：

依次找出丑数，从最小的丑数开始用K个素数来扩增，如果这些扩曾出的数字P，没有被加入队列，那么就加入队列。 扩增完后，当前最小丑数出队。 从下一个丑数开始扩增……

我用了SPLAY，我把队列大小限制在50W以内……然后第N个出队的才是正确答案。 40W都不行…… 如果不限制队列大小，貌似计算太慢太慢…… 只能过11个测试点。

Compiling...Compile: OKExecuting...   Test 1: TEST OK [0.003 secs, 3504 KB]   Test 2: TEST OK [0.003 secs, 3504 KB]   Test 3: TEST OK [0.005 secs, 3504 KB]   Test 4: TEST OK [0.032 secs, 3636 KB]   Test 5: TEST OK [0.049 secs, 3768 KB]   Test 6: TEST OK [0.211 secs, 4692 KB]   Test 7: TEST OK [0.084 secs, 4032 KB]   Test 8: TEST OK [0.062 secs, 3900 KB]   Test 9: TEST OK [0.003 secs, 3504 KB]   Test 10: TEST OK [0.003 secs, 3504 KB]   Test 11: TEST OK [0.003 secs, 3504 KB]

  > Run 12: Execution error:

/*TASK:humbleLANG:C++*/#include <cstdio> int k, n;int a[102];void init(){scanf("%d%d", &k, &n);for (int i = 0; i != k; ++ i)scanf("%d", &a[i]);} struct node{node *c[2];int key, size;node(){key = size = 0;c[0] = c[1] = this;}node(int KEY_, node *c0, node *c1){key = KEY_;c[0] = c0;c[1] = c1;}node* rz(){return size = c[0] -> size + c[1] -> size + 1, this;}}Tnull, *null = &Tnull; const int maxint = 0x7fffffff;struct splay{node *root;splay(){root = (new node(*null)) -> rz();root -> key = maxint;} void zig(bool d){node *t = root -> c[d];root -> c[d] = null -> c[d];null -> c[d] = root;root = t;} void zigzig(bool d){node *t = root -> c[d] -> c[d];root -> c[d] -> c[d] = null -> c[d];null -> c[d] = root -> c[d];root -> c[d] = null -> c[d] -> c[!d];null -> c[d] -> c[!d] = root -> rz();root = t;} void finish(bool d){node *t = null -> c[d], *p = root -> c[!d];while (t != null){t = null -> c[d] -> c[d];null -> c[d] -> c[d] = p;p = null -> c[d] -> rz();null -> c[d] = t;}root -> c[!d] = p;} void select(int k)//谁有k个儿子{int t;while (1){bool d = k > (t = root -> c[0] -> size);if (k == t || root -> c[d] == null)break;if (d)k -= t + 1;bool dd = k > (t = root -> c[d] -> c[0] -> size);if (k == t || root -> c[d] -> c[dd] == null){zig(d);break;}if (dd)k -= t + 1;d != dd ? zig(d), zig(dd) : zigzig(d);}finish(0), finish(1);root -> rz();} void search(int x){while (1){bool d = x > root -> key;if (root -> c[d] == null)break;bool dd = x > root -> c[d] -> key;if (root -> c[d] -> c[dd] == null){zig(d); break;}d != dd ? zig(d), zig(dd) : zigzig(d);}finish(0), finish(1);root -> rz();if (x > root -> key)select(root -> c[0] -> size + 1);} void ins(int x){search(x);node *oldroot = root;root = new node(x, oldroot -> c[0], oldroot);oldroot -> c[0] = null;oldroot -> rz();root -> rz();}void del(int x){search(x);node *oldroot = root;root = root -> c[1];select(0);root -> c[0] = oldroot -> c[0];root -> rz();delete oldroot;}int sel(int k){return select(k - 1), root -> key;}int ran(int x){return search(x), root -> c[0] -> size + 1;}}sp;  void doit(){sp.ins(1);int count = -1;while (1){int now = sp.sel(1);++ count;if (count == n){printf("%d\n",now);return;}if (sp.root -> size >= 2000000)continue;for (int i = 0; i != k; ++ i){long long tmp = (long long)now * a[i];if (tmp > 2000000000)continue;sp.search(tmp);int  p = sp.root -> key;if (sp.root->key == tmp)continue;else sp.ins(tmp);}now = sp.sel(1);sp.del(now);}} int main(){freopen("humble.in","r",stdin);freopen("humble.out","w",stdout);init();doit();return 0;}

正确方法：

思路： 假如已经有P个丑数，那么我可以把这P个丑数，分别和K个已知素数进行相乘，得到的最小的（同时不能是已经得到的丑数），就是第P+1个丑数。

这其中有一个搜索优化。

第i个素数， 假如和第j个丑数，产生了最小的，也就是第p+1个丑数。 那么这第i个素数，以后如果要产生第X个丑数，他一定从第j个丑数之后搜索。【有点拗口】

细节： 从丑数5，可以产生5*2 = 10.  从丑数2， 可以产生2*5 = 10.  所以在进行产生丑数运算前，要先保证，所有的第i个素数，和他们“当前搜索丑是数j” ，保证素数i * 丑数j > 第P个丑数。

Executing...   Test 1: TEST OK [0.003 secs, 3756 KB]   Test 2: TEST OK [0.003 secs, 3756 KB]   Test 3: TEST OK [0.003 secs, 3756 KB]   Test 4: TEST OK [0.005 secs, 3756 KB]   Test 5: TEST OK [0.008 secs, 3756 KB]   Test 6: TEST OK [0.022 secs, 3756 KB]   Test 7: TEST OK [0.005 secs, 3756 KB]   Test 8: TEST OK [0.005 secs, 3756 KB]   Test 9: TEST OK [0.003 secs, 3756 KB]   Test 10: TEST OK [0.003 secs, 3756 KB]   Test 11: TEST OK [0.003 secs, 3756 KB]   Test 12: TEST OK [0.081 secs, 3756 KB]All tests OK.

/*TASK:humbleLANG:C++*/#include <cstdio>int k, n;int a[102], w[102]={0}, x[100005] = {1};int main(){freopen("humble.in","r",stdin);freopen("humble.out","w",stdout);scanf("%d%d", &k, &n);for (int i = 0; i != k; ++ i)scanf("%d", &a[i]);for (int j = 1; j <= n; ++ j){int tmp = 0x7fffffff, wz;for (int i = 0; i != k; ++ i)while (x[w[i]] * a[i] <= x[j - 1])++ w[i];for (int i = 0; i != k; ++ i){if (x[w[i]] * a[i] < tmp){tmp = x[w[i]] * a[i];wz = i;}}++ w[wz];x[j] = tmp;}printf("%d\n", x[n]);return 0;}