2017 ACM-ICPC 亞洲區(南寧賽區)網絡賽 A Weather Patterns

Consider a system which is described at any time as being in one of a set of NN distinct states, 1,2,3,...,N1,2,3,...,N. We denote the time instants associated with state changes as t = 1,2,...t=1,2,..., and the actual state at time tt as a_{ij}=p=[s_{i}=j\ |\ s_{i-1}=i], 1\le i,j \le Naij=p=[si=j  si1=i],1i,jN. For the special case of a discrete, first order, Markovchain, the probabilistic description for the current state (at time tt) and the predecessor state is s_{t}st. Furthermore we only consider those processes being independent of time, thereby leading to the set of state transition probability a_{ij}aij of the form: with the properties a_{ij} \geq 0aij0 and \sum_{i=1}^{N} A_{ij} = 1i=1NAij=1. The stochastic process can be called an observable Markovmodel. Now, let us consider the problem of a simple 4-state Markov model of weather. We assume that once a day (e.g., at noon), the weather is observed as being one of the following:

State 11: snow

State 22: rain

State 33: cloudy

State 44: sunny

The matrix A of state transition probabilities is:

A = \{a_{ij}\}= \begin{Bmatrix} a_{11}&a_{12}&a_{13}&a_{14} \\ a_{21}&a_{22}&a_{23}&a_{24} \\ a_{31}&a_{32}&a_{33}&a_{34} \\ a_{41}&a_{42}&a_{43}&a_{44} \end{Bmatrix}A={aij}=a11a21a31a41a12a22a32a42a13a23a33a43a14a24a34a44

Given the model, several interestingquestions about weather patterns over time can be asked (and answered). We canask the question: what is the probability (according to the given model) thatthe weather for the next k days willbe? Another interesting question we can ask: given that the model is in a knownstate, what is the expected number of consecutive days to stay in that state?Let us define the observation sequence OOas O = \left \{ s_{1}, s_{2}, s_{3}, ... , s_{k} \right \}O={s1,s2,s3,...,sk}, and the probability of the observation sequence OOgiven the model is defined as p(O|model)p(Omodel). Also, let the expected number of consecutive days to stayin state ii be E_{i}Ei. Assume that the initial state probabilities p[s_{1} = i] = 1, 1 \leq i \leq Np[s1=i]=1,1iN. Bothp(O|model)p(Omodel)and E_{i}Ei are real numbers.

Input Format

Line 11~44 for the state transition probabilities. Line 55 for the observation sequence O_{1}O1, and line 66 for the observation sequence O_{2}O2. Line 77and line 88 for the states of interest to find the expected number of consecutive days to stay in these states.

Line 11a_{11}\ a_{12}\ a_{13}\ a_{14}a11 a12 a13 a14

Line 22a_{21}\ a_{22}\ a_{23}\ a_{34}a21 a22 a23 a34

Line 33a_{31}\ a_{32}\ a_{33}\ a_{34}a31 a32 a33 a34

Line 44a_{41}\ a_{42}\ a_{43}\ a_{44}a41 a42 a43 a44

Line 55s_{1}\ s_{2}\ s_{3}\ ...\ s_{k}s1 s2 s3 ... sk

Line 66s_{1}\ s_{2}\ s_{3}\ ...\ s_{l}s1 s2 s3 ... sl

Line 77ii

Line 88jj

Output Format

Line 11 and line 22 are used to show the probabilities of the observation sequences O_{1}O1and O_{2}O2 respectively. Line 33 and line 44 are for the expected number of consecutive days to stay in states ii and jj respectively.

Line 11p[O_{1} | model]p[O1model]

Line 22p[O_{2} | model]p[O2model]

Line 33E_{i}Ei

Line 44E_{j}Ej

Please be reminded that the floating number should accurate to 10^{-8}108.

樣例輸入

0.4 0.3 0.2 0.1
0.3 0.3 0.3 0.1
0.1 0.1 0.6 0.2
0.1 0.2 0.2 0.5
4 4 3 2 2 1 1 3 3
2 1 1 1 3 3 4
3
4

樣例輸出

0.00004320
0.00115200
2.50000000
2.00000000

題目來源

2017 ACM-ICPC 亞洲區(南寧賽區)網絡賽

題意:先給你4 * 4 的矩陣,表示天氣 i 到天氣 j 轉變的概率,然後輸入 k 個數字,以 ‘\n’ 表示輸入結束,這 k 個數字的意思以
4 4 3 2 2 1 1 3 3

這組數字舉例,其實就是天氣從 4 轉變爲 4,然後從 4 轉變爲 3,然後從 3 轉變爲 2,然後從 2 轉變爲 2,然後從 2 轉變爲 1,然後從 1 轉變爲 1,然後從1 轉變爲 3,然後從3 轉變爲 3的概率是多少。

這樣的輸入數據有兩組

然後輸入兩個數字 i 和 j ,這兩個數字都表示天氣的狀況爲 i 和 j 一直不變,不管天數增加多少,然後要你求一直不變的數學期望是多少。



附代碼:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>
#define mem(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long LL;
///o(っ。Д。)っ AC萬歲!!!!!!!!!!!!!!
const int maxn = 10;
double maps[maxn][maxn] = {};


int main()
{
    for(int i = 1; i <= 4; i++)
    {
        for(int j = 1; j <= 4; j++)
        {
            scanf("%lf", &maps[i][j]);
        }
    }
    double pp = 1.0;
    int d1, d2;
    char ch;
    scanf("%d", &d1);
    while(scanf("%d%c", &d2, &ch) != EOF)
    {
        pp *= maps[d1][d2];
        d1 = d2;
        if(ch == '\n') break;
    }
    printf("%.8lf\n", pp);
    pp = 1.0;
    scanf("%d", &d1);
    while(scanf("%d%c", &d2, &ch) != EOF)
    {
        pp *= maps[d1][d2];
        d1 = d2;
        if(ch == '\n') break;
    }
    printf("%.8lf\n", pp);
    scanf("%d", &d1);
    pp = maps[d1][d1];
    double sum = 0.0;
    while(pp >= 0.000000001)
    {
        sum += pp;
        pp *= maps[d1][d1];
    }
    printf("%.8lf\n", sum + 1.0);
    sum = 0.0;
    scanf("%d", &d2);
    pp = maps[d2][d2];
    while(pp >= 0.000000001)
    {
        sum += pp;
        pp *= maps[d2][d2];
    }
    printf("%.8lf\n", sum + 1.0);
    return 0;
}
/*


*/





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