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

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
double dp[10][130], mat[130][130];
int n;

int main()
{
    int i, j, k, t;
    while(scanf("%d", &n) && (n != -1)) {
        for(i = 1; i <= (1 << n); ++i)
            for(j = 1; j <= (1 << n); ++j)
                scanf("%lf", &mat[i][j]);
        memset(dp, 0, sizeof(dp));
        int tot = 1 << n, half;
        for(i = 1; i <= tot; ++i)
            dp[0][i] = 1;
        for(i = 1; i <= n; ++i) {
            tot = 1 << i, half = 1 << (i - 1);
            for(j = 1; j <= (1 << n); j += tot) {
                for(k = j; k <= j + half - 1; ++k) {
                    for(t = j + half; t <= j + tot - 1; ++t) {
                        dp[i][k] += dp[i - 1][k] * dp[i - 1][t] * mat[k][t];
                        dp[i][t] += dp[i - 1][k] * dp[i - 1][t] * mat[t][k];
                    }
                }
            }
        }
        double ans = 0;
        int res = 1;
        for(i = 1; i <= (1 << n); ++i)
            if(ans < dp[n][i])
                ans = dp[n][i], res = i;
        printf("%d
", res);
    }
}

//#2 比较向上 i - 1 层的根节点序号（存在提前相遇情况）

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
double dp[10][130], mat[130][130];
int n;

int main()
{
    int i, j, k;
    while(scanf("%d", &n) && (n != -1)) {
        for(i = 1; i <= (1 << n); ++i)
            for(j = 1; j <= (1 << n); ++j)
                scanf("%lf", &mat[i][j]);
        memset(dp, 0, sizeof(dp));
        for(i = 1; i <= (1 << n); ++i)
            dp[0][i] = 1;
        for(i = 1; i <= n; ++i) {
            for(j = 1; j <= (1 << n); ++j) {
                for(k = 1; k <= (1 << n); ++k) {
                    int root_j = (j - 1) >> (i - 1), root_k = (k - 1) >> (i - 1);
                    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
                        dp[i][j] += dp[i - 1][j] * dp[i - 1][k] * mat[j][k];
                        dp[i][k] += dp[i - 1][j] * dp[i - 1][k] * mat[k][j];
                    }
                }
            }
        }
        int res = 1;
        double ans = 0;
        for(i = 1; i <= (1 << n); ++i)
            if(ans < dp[n][i])
                ans = dp[n][i], res = i;
        printf("%d
", res);
    }
}