poj 2151 Check the difficulty of problems(线段树+概率）

标签：树形结构-线段树

Description

Input

Output

Sample Input

2 2 2
0.9 0.9
1 0.9
0 0 0

Sample Output

0.972

每个节点记录这个区间每个队赢的概率。

#include <iostream>
#include <cstdio>
using namespace std;

const int maxn = 200;
const double eps = 0.0001;
struct tree{
    int l , r;
    double win[maxn];
}a[4*maxn];
double P[maxn][maxn];
int N[10] = {1 , 2 , 4 , 8 , 16 , 32 , 64 , 128 , 256 , 512};
int n;

void initial(){
    for(int i = 0; i < 4*maxn; i++){
        for(int j = 0; j < maxn; j++){
            a[i].win[j] = 0.0;
        }
    }
}

void build(int l , int r , int k){
    a[k].l = l;
    a[k].r = r;
    if(l == r){
        a[k].win[l] = 1.0;
    }else{
        int mid = (l+r)/2;
        build(l , mid , 2*k);
        build(mid+1 , r , 2*k+1);
        int i = l;
        while(i <= mid){
            for(int j = mid+1; j <= r; j++){
                a[k].win[i] += a[2*k].win[i]*a[2*k+1].win[j]*P[i][j];
            }
            i++;
        }
        while(i <= r){
            for(int j = l; j <= mid; j++){
                a[k].win[i] += a[2*k].win[j]*a[2*k+1].win[i]*P[i][j];
            }
            i++;
        }
    }
}

void readcase(){
    for(int i = 0; i < N[n]; i++){
        for(int j = 0; j < N[n]; j++){
            scanf("%lf" , &P[i][j]);
        }
    }
}

void computing(){
    build(0 , N[n]-1 , 1);
    int ans = 0;
    for(int i = 1; i < N[n]; i++){
        if(a[1].win[i] - a[1].win[ans] > eps){
            ans = i;
        }
    }
    printf("%d\n" , ans+1);
}

int main(){
    while(scanf("%d" , &n) && n != -1){
        initial();
        readcase();
        computing();
    }
    return 0;
}

poj 2151 Check the difficulty of problems(线段树+概率）,布布扣,bubuko.com

poj 2151 Check the difficulty of problems(线段树+概率）

标签：树形结构-线段树

原文地址：http://blog.csdn.net/u011836218/article/details/32931701