POJ3071 Football

嘟嘟嘟


这题还给样例解释了，那自然能想到dp。
因为比赛形式构成了一个树形结构，而且还和线段树的结构一模一样，那就索性这么dp：令\(dp[u][i]\)表示在节点\(u\)，\(i\)获胜的概率。
然后我们枚举左右儿子谁赢，就很容易搞出转移方程：$$dp[u][i] = \sum dp[u << 1][i] * dp[u << 1 | 1][j] * p[i][j] \dp[u][i] = \sum dp[u << 1 | 1][i] * dp[u << 1][j] * p[i][j]$$。
当递归到叶子节点的时候，初始化\(dp[u][L] = 1\)。至于怎么判断叶子节点，就跟线段树一样。


看了很多题解都找到了一个什么规律，反正找规律方面我的水平是0……

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
#include<assert.h>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = (1 << 7) + 5;
In ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
In void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
  freopen(".in", "r", stdin);
  freopen(".out", "w", stdout);
#endif
}

int K, n;
db p[maxn][maxn], dp[maxn << 2][maxn];

In void dfs(int now, int L, int R)
{
  for(int i = L; i <= R; ++i) dp[now][i] = 0;
  if(L == R) {dp[now][L] = 1; return;}
  int mid = (L + R) >> 1;
  dfs(now << 1, L, mid), dfs(now << 1 | 1, mid + 1, R);
  for(int i = L; i <= mid; ++i)
    for(int j = mid + 1; j <= R; ++j)
      dp[now][i] += dp[now << 1][i] * dp[now << 1 | 1][j] * p[i][j];
  for(int i = mid + 1; i <= R; ++i)
    for(int j = L; j <= mid; ++j)
      dp[now][i] += dp[now << 1 | 1][i] * dp[now << 1][j] * p[i][j];
}

int main()
{
  //MYFILE();
  while(scanf("%d", &K) && ~K)
    {
      n = 1 << K;
      for(int i = 1; i <= n; ++i)
	for(int j = 1; j <= n; ++j) scanf("%lf", &p[i][j]);
      dfs(1, 1, n);
      int pos = 1;
      for(int i = 2; i <= n; ++i) if(dp[1][i] > dp[1][pos]) pos = i;
      write(pos), enter;
    }
  return 0;
}