uva 10247 - Complete Tree Labeling（dp）

题目链接：uva 10247 - Complete Tree Labeling

题目大意：给出k和d，表示有一个k叉d层的完全k叉数，然后它的节点数为n个，用1~n给这棵树的节点标号，要求说任意一个节点的值不能大于它的任意一个子节点，每个数只能用一次。问优多少种标记的方法。

解题思路：一般dp都是有dp[i - 1]推导出dp[i]的，这题也不例外，只不过思路和往常不太一样。节点数node[i][j]表示说完全i叉数j层树含有几个节点，node[i][j] = node[i][j - 1] * i + 1,然后ans[i][j] 即为i叉树的标记方式数。然后对于每个ans[i][j]可以看成是i个ans[i][j - 1]（子树），根节点肯定是1.然后对于每颗子树需要m = node[i][j - 1]个数来标记节点（数为有序的，并且不相等），就要ans[i][j] = C(node[i][j] - 1, m) * C(node[i][j] - 1 - m, m) * ....* C(m, m) * (ans[i][j - 1] ^ i).

#include <stdio.h>#include <string.h>#include <math.h>#define max(a,b) (a)>(b)?(a):(b)#define min(a,b) (a)<(b)?(a):(b)const int MAXSIZE = 10000;struct bign {int s[MAXSIZE];bign (){memset(s, 0, sizeof(s));}bign (int number) {*this = number;}bign (const char* number) {*this = number;}    void put();bign mul(int d);void del();void init() { memset(s, 0, sizeof(s)); }    bign operator =  (char *num);bign operator =  (int num);bool operator <  (const bign& b) const;bool operator >  (const bign& b) const { return b < *this; }bool operator <= (const bign& b) const { return !(b < *this); }bool operator >= (const bign& b) const { return !(*this < b); }bool operator != (const bign& b) const { return b < *this || *this < b;}bool operator == (const bign& b) const { return !(b != *this); }    bign operator + (const bign& c);bign operator * (const bign& c);bign operator - (const bign& c);int  operator / (const bign& c);bign operator / (int k);bign operator % (const bign &c);int  operator % (int k);void operator ++ ();bool operator -- ();};const int N = 22;int node[N][N];bign ans[N][N];bign C(int x, int y) {y = min(y, x - y);bign c = 1, d;for (int i = 0; i < y; i++) {d = x - i;c = c * d;c = c / (i + 1);}return c;}void solve(int x, int y) {ans[x][y] = 1;int m = node[x][y - 1];for (int i = 0; i < x; i++) {ans[x][y] = ans[x][y] * C(node[x][y] - i * m - 1, m) * ans[x][y - 1];}}void init() {for (int i = 1; i <= 21; i++)ans[1][i] = 1;for (int i = 2; i <= 21; i++) {int top = 21 / i;ans[i][0] = node[i][0] = 1;node[i][0] = 1;for (int j = 1; j <= top; j++) {node[i][j] = node[i][j - 1] * i + 1;solve(i, j);}}}int main () {init();int k, d;while (scanf("%d%d", &k, &d) == 2) {ans[k][d].put();printf("\n");}/*int a, b;while (scanf("%d%d", &a, &b) == 2) {bign t = C(a, b);t.put();printf("!\n");}*/return 0;}bign bign::operator = (char *num) {init();s[0] = strlen(num);for (int i = 1; i <= s[0]; i++)s[i] = num[s[0] - i] - '0';return *this;}bign bign::operator = (int num) {char str[MAXSIZE];sprintf(str, "%d", num);return *this = str;}bool bign::operator < (const bign& b) const {if (s[0] != b.s[0])return s[0] < b.s[0];for (int i = s[0]; i; i--)if (s[i] != b.s[i])return s[i] < b.s[i];return false;}bign bign::operator + (const bign& c) {int sum = 0;bign ans;ans.s[0] = max(s[0], c.s[0]);for (int i = 1; i <= ans.s[0]; i++) {if (i <= s[0]) sum += s[i];if (i <= c.s[0]) sum += c.s[i];ans.s[i] = sum % 10;sum /= 10;}return ans;}bign bign::operator * (const bign& c) {bign ans;ans.s[0] = 0; for (int i = 1; i <= c.s[0]; i++){  int g = 0;  for (int j = 1; j <= s[0]; j++){  int x = s[j] * c.s[i] + g + ans.s[i + j - 1];  ans.s[i + j - 1] = x % 10;  g = x / 10;  }  int t = i + s[0] - 1;while (g){  ++t;g += ans.s[t];ans.s[t] = g % 10;g = g / 10;  }  ans.s[0] = max(ans.s[0], t);}  ans.del();return ans;}bign bign::operator - (const bign& c) {bign ans = *this;for (int i = 1; i <= c.s[0]; i++) {if (ans.s[i] < c.s[i]) {ans.s[i] += 10;ans.s[i + 1]--;;}ans.s[i] -= c.s[i];}for (int i = 1; i <= ans.s[0]; i++) {if (ans.s[i] < 0) {ans.s[i] += 10;ans.s[i + 1]--;}}ans.del();return ans;}int bign::operator / (const bign& c) {int ans = 0;bign d = *this;while (d >= c) {d = d - c;ans++;}return ans;}bign bign::operator / (int k) {bign ans; ans.s[0] = s[0];int num = 0;  for (int i = s[0]; i; i--) {  num = num * 10 + s[i];  ans.s[i] = num / k;  num = num % k;  }  ans.del();return ans;}int bign:: operator % (int k){  int sum = 0;  for (int i = s[0]; i; i--){  sum = sum * 10 + s[i];  sum = sum % k;  }  return sum;  } bign bign::operator % (const bign &c) {bign now = *this;while (now >= c) {now = now - c;now.del();}return now;}void bign::operator ++ () {s[1]++;for (int i = 1; s[i] == 10; i++) {s[i] = 0;s[i + 1]++;s[0] = max(s[0], i + 1);}}bool bign::operator -- () {del();if (s[0] == 1 && s[1] == 0) return false;int i;for (i = 1; s[i] == 0; i++)s[i] = 9;s[i]--;del();return true;}void bign::put() {if (s[0] == 0)printf("0");elsefor (int i = s[0]; i; i--)printf("%d", s[i]);}bign bign::mul(int d) {s[0] += d;for (int i = s[0]; i > d; i--)s[i] = s[i - d];for (int i = d; i; i--)s[i] = 0;return *this;}void bign::del() {while (s[s[0]] == 0) {s[0]--;if (s[0] == 0) break;}}