标签:style class code ext width color string int html set htm
给定2^n 支足球队进行比赛,n<=7. 队伍两两之间有一个获胜的概率,求每一个队伍赢得最后比赛的概率是多少?
状态其实都是很显然的,一开始觉得这个问题很难啊,不会。dp[i][j] 表示第i支队伍赢得前j轮比赛的概率。(这个题目处理区间的时候比较恶心,小心点即可)。
1: 2: #include <iostream> 3: #include <cstdio> 4: #include <cstring> 5: #include <map> 6: #include <algorithm>7: using namespace std;
8: 9: double p[135][135];
10: double dp[135][10];
11: 12: pair<int,int> range(int idx, int round)
13: {14: int x = (idx>>round) ^1;
15: return make_pair(x<<round, (x<<round) + (1<<(round)) - 1);
16: }17: int main()
18: {19: // freopen("1.txt","r",stdin);
20: int n;
21: while(cin>>n && n !=-1)
22: {23: for(int i=0; i<(1<<n); i++)
24: {25: for(int j=0; j<(1<<n); j++)
26: { 27: cin>>p[i][j]; 28: } 29: }30: memset(dp,0,sizeof(dp));
31: for(int i=0; i< (1<<n); i++)
32: { 33: dp[i][0] = p[i][i^1]; 34: }35: for(int round=1; round<n; round++) // round
36: {37: for(int j=0; j< (1<<n); j++)
38: {39: pair<int,int> r = range(j, round);
40: for(int k = r.first; k<= r.second; k++)
41: { 42: dp[j][round] += p[j][k] *dp[k][round - 1]*dp[j][round-1]; 43: } 44: } 45: }46: int idx = -1;
47: double ret = 0;
48: double sum = 0;
49: for(int i=0; i<(1<<n); i++)
50: { 51: sum+= dp[i][n-1];52: if(dp[i][n-1] > ret)
53: { 54: idx = i; 55: ret = dp[i][n-1]; 56: } 57: } 58: cout<<idx+1<<endl; 59: } 60: }标签:style class code ext width color string int html set htm
原文地址:http://www.cnblogs.com/sosi/p/3701410.html