【POJ 3071】 Football（DP）
【POJ 3071】 Football（DP）
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 4350 | Accepted: 2222 |
Description
Consider a single-elimination football tournament involving 2n teams, denoted 1, 2, …, 2n. In each round of the tournament, all teams still in the tournament are placed in a list in order of increasing index. Then, the first team in the list plays the second team, the third team plays the fourth team, etc. The winners of these matches advance to the next round, and the losers are eliminated. After n rounds, only one team remains undefeated; this team is declared the winner.
Given a matrix P = [pij] such that pij is the probability that team i will beat team j in a match determine which team is most likely to win the tournament.
Input
The input test file will contain multiple test cases. Each test case will begin with a single line containing n (1 ≤ n ≤ 7). The next 2n lines each contain 2n values; here, the jth value on the ith line represents pij. The matrix P will satisfy the constraints that pij = 1.0 − pji for all i ≠ j, and pii = 0.0 for all i. The end-of-file is denoted by a single line containing the number −1. Note that each of the matrix entries in this problem is given as a floating-point value. To avoid precision problems, make sure that you use either the double data type instead of float.
Output
The output file should contain a single line for each test case indicating the number of the team most likely to win. To prevent floating-point precision issues, it is guaranteed that the difference in win probability for the top two teams will be at least 0.01.
Sample Input
2 0.0 0.1 0.2 0.3 0.9 0.0 0.4 0.5 0.8 0.6 0.0 0.6 0.7 0.5 0.4 0.0 -1
Sample Output
2
Hint
In the test case above, teams 1 and 2 and teams 3 and 4 play against each other in the first round; the winners of each match then play to determine the winner of the tournament. The probability that team 2 wins the tournament in this case is:
| P(2 wins) | = P(2 beats 1)P(3 beats 4)P(2 beats 3) + P(2 beats 1)P(4 beats 3)P(2 beats 4) = p21p34p23 + p21p43p24 = 0.9 · 0.6 · 0.4 + 0.9 · 0.4 · 0.5 = 0.396. |
The next most likely team to win is team 3, with a 0.372 probability of winning the tournament.
算是个概率dp，。。比较简单的
题目大意:要举办n场足球比赛，一共有2^n支队伍。
比赛规则就是晋级型，第一个跟第二个比，第三个跟第四个，每场中赢的一支队伍进入下一场比赛。
最后只有一个冠军。
大体就是树型的那种。
问有最大概率获得冠军的队伍编号，题目还保证不会有精度问题了。
这样n <= 7 最多1<<7 = 128个队伍。
dp[i][j] 表示编号为i的队伍在第j场比赛中胜出的概率
由于是树型，其实当前场次每组胜出的队伍就是这个子树的根，他会与同父亲的另一棵子树，或者说和他兄弟中的队伍比赛。
这样每次暴力枚举赢家，然后求出在该场胜出的概率即可
代码如下:
#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define Pr pair<int,int>
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)
using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const int mod = 1e9+7;
const double eps = 1e-8;
double win[133][133];
double dp[133][8];
int main()
{
//fread();
//fwrite();
int n,m;
while(~scanf("%d",&n) && ~n)
{
m = n;
memset(dp,0,sizeof(dp));
n = 1<<n;
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j)
scanf("%lf",&win[i][j]);
int ad,st,en;
for(int i = 0; i < m; ++i)
{
//当前场次覆盖的区间范围大小
ad = 1<<i;
//printf("level:%d can:%d mx:%d ad:%d\n",i,tmp,k,ad);
//st-ad表示当前球队所在范围
int st = 1, en = ad;
for(int j = 1; j <= n; ++j)
{
if(!i)
{
//printf("%dto%d\n",j,j+(j&1? 1: -1));
dp[j][i] = win[j][j+(j&1? 1: -1)];
}
else
{
if(j > en)
{
st += ad;
en += ad;
}
//printf("%d-%d\n",st,en);
//左子树
if(en&ad)
{
//printf("findin:%d-%d\n",st+ad,en+ad);
for(int z = st+ad; z <= en+ad; ++z)
dp[j][i] += win[j][z]*dp[z][i-1];
}
//右子树
else
{
//printf("findin:%d-%d\n",st-ad,en-ad);
for(int z = st-ad; z <= en-ad; ++z)
dp[j][i] += win[j][z]*dp[z][i-1];
}
dp[j][i] *= dp[j][i-1];
}
}
}
int id = 1;
for(int i = 2; i <= n; ++i)
{
//printf("%d %f\n",i,dp[i][m-1]);
if(dp[id][m-1] < dp[i][m-1]) id = i;
}
printf("%d\n",id);
}
return 0;
}