poj 3071 概率dp
//分析：明显的树形关系，题目描述的是一棵高 n + 1的完全二叉树，则 dp[树层号][team号]（规定最底层为 0 层，层数朝节点的方向依次递增），推一下就好了
//稍微需要想一下的是比赛双方的选取，下面给出两种方法
//#1 枚举起点划分team区间
1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 double dp[10][130], mat[130][130]; 7 int n; 8 9 int main() 10 { 11 int i, j, k, t; 12 while(scanf("%d", &n) && (n != -1)) { 13 for(i = 1; i <= (1 << n); ++i) 14 for(j = 1; j <= (1 << n); ++j) 15 scanf("%lf", &mat[i][j]); 16 memset(dp, 0, sizeof(dp)); 17 int tot = 1 << n, half; 18 for(i = 1; i <= tot; ++i) 19 dp[0][i] = 1; 20 for(i = 1; i <= n; ++i) { 21 tot = 1 << i, half = 1 << (i - 1); 22 for(j = 1; j <= (1 << n); j += tot) { 23 for(k = j; k <= j + half - 1; ++k) { 24 for(t = j + half; t <= j + tot - 1; ++t) { 25 dp[i][k] += dp[i - 1][k] * dp[i - 1][t] * mat[k][t]; 26 dp[i][t] += dp[i - 1][k] * dp[i - 1][t] * mat[t][k]; 27 } 28 } 29 } 30 } 31 double ans = 0; 32 int res = 1; 33 for(i = 1; i <= (1 << n); ++i) 34 if(ans < dp[n][i]) 35 ans = dp[n][i], res = i; 36 printf("%d\n", res); 37 } 38 }
//#2 比较向上 i - 1 层的根节点序号（存在提前相遇情况）
1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 double dp[10][130], mat[130][130]; 7 int n; 8 9 int main() 10 { 11 int i, j, k; 12 while(scanf("%d", &n) && (n != -1)) { 13 for(i = 1; i <= (1 << n); ++i) 14 for(j = 1; j <= (1 << n); ++j) 15 scanf("%lf", &mat[i][j]); 16 memset(dp, 0, sizeof(dp)); 17 for(i = 1; i <= (1 << n); ++i) 18 dp[0][i] = 1; 19 for(i = 1; i <= n; ++i) { 20 for(j = 1; j <= (1 << n); ++j) { 21 for(k = 1; k <= (1 << n); ++k) { 22 int root_j = (j - 1) >> (i - 1), root_k = (k - 1) >> (i - 1); 23 if(root_j != root_k && (root_j >> 1) == (root_k >> 1)) { //向上 i - 1 层的根节点不同 且 是同一节点的左右孩子 24 dp[i][j] += dp[i - 1][j] * dp[i - 1][k] * mat[j][k]; 25 dp[i][k] += dp[i - 1][j] * dp[i - 1][k] * mat[k][j]; 26 } 27 } 28 } 29 } 30 int res = 1; 31 double ans = 0; 32 for(i = 1; i <= (1 << n); ++i) 33 if(ans < dp[n][i]) 34 ans = dp[n][i], res = i; 35 printf("%d\n", res); 36 } 37 }
