POJ-3744 Scout YYF I （概率DP&&矩阵快速幂）
Scout YYF I
http://poj.org/problem?id=3744
| Time Limit: 1000MS | Memory Limit: 65536K | |
Description
YYF is a couragous scout. Now he is on a dangerous mission which is to penetrate into the enemy's base. After overcoming a series difficulties, YYF is now at the start of enemy's famous "mine road". This is a very long road, on which there are numbers of mines. At first, YYF is at step one. For each step after that, YYF will walk one step with a probability of
p, or jump two step with a probality of 1-
p. Here is the task, given the place of each mine, please calculate the probality that YYF can go through the "mine road" safely.
Input
The input contains many test cases ended with
EOF.
Each test case contains two lines.
The First line of each test case is N (1 ≤ N ≤ 10) and p (0.25 ≤ p ≤ 0.75) seperated by a single blank, standing for the number of mines and the probability to walk one step.
The Second line of each test case is N integer standing for the place of N mines. Each integer is in the range of [1, 100000000].
Each test case contains two lines.
The First line of each test case is N (1 ≤ N ≤ 10) and p (0.25 ≤ p ≤ 0.75) seperated by a single blank, standing for the number of mines and the probability to walk one step.
The Second line of each test case is N integer standing for the place of N mines. Each integer is in the range of [1, 100000000].
Output
For each test case, output the probabilty in a single line with the precision to 7 digits after the decimal point.
Sample Input
1 0.5 2 2 0.5 2 4
Sample Output
0.5000000 0.2500000
题目大意：在一条很长的路上，有n个雷，给出每个雷的坐标，以及一个人每次走一步的概率p，走两步的概率为1-p，初始在点1处，求安全走过这段路的概率？
大致思路：设dp[i]表示安全到达点i处的概率，则状态转移方程为：
①点i处有雷，dp[i]=0;
②点i出无雷，dp[i]=dp[i-1]*p+dp[i-2]*(1-p);
由于雷坐标的范围很大，而雷的数量很少，所以需要用矩阵快速幂对无雷的路段进行优化
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN=2;
int n,index[13];
double p,ans;
struct Matrix{
double m[MAXN][MAXN];
}ORI;
Matrix I = {1,0,
0,1};
Matrix matrixmul(const Matrix& a,const Matrix& b) {//矩阵乘法
int i,j,k;
Matrix c;
for (i = 0 ; i < MAXN; i++)
for (j = 0; j < MAXN;j++) {
c.m[i][j] = 0;
for (k = 0; k < MAXN; k++)
c.m[i][j] += a.m[i][k] * b.m[k][j];
}
return c;
}
Matrix quickpow(int n) {
Matrix m = ORI, b = I;
while (n >= 1) {
if (n & 1)
b = matrixmul(b,m);
n = n >> 1;
m = matrixmul(m,m);
}
return b;
}
int main() {
while(2==scanf("%d%lf",&n,&p)) {
ORI.m[0][0]=p;
ORI.m[0][1]=1;
ORI.m[1][0]=1-p;
ORI.m[1][1]=0;
for(int i=1;i<=n;++i) {
scanf("%d",index+i);
}
index[0]=0;
sort(index,index+n+1);
ans=1;//ans表示到达index[i]+1处的概率
for(int i=1;i<=n;++i) {
if(index[i]!=index[i-1]) {//如果两个雷不在同一点
if(index[i]-index[i-1]==1) {//如果两个雷相邻，则无法安全通过
ans=0.00000001;
break;
}
ans*=quickpow(index[i]-index[i-1]-2).m[0][0];//算出安全到达index[i]-1的概率
ans*=1-p;//算出安全到达index[i]+1的概率
}
}
printf("%.7lf\n",ans);
}
return 0;
}
for(int i=1;i<=n;++i) {
if(index[i]!=index[i-1]) {//如果两个雷不在同一点
ans*=(1-quickpow(index[i]-index[i-1]-1).m[0][0]);//直接算出安全到达index[i]+1的概率
}
}
表示不明白如何推出来（只会将p带入方程，计算推出n较小时成立），估计很简单，大家都没有说明是怎么来的...
终于还是找到一个差不多的解释：将 index[i-1]+1~index[i]-1看成一个点，则下一步只可能到达index[i]或index[i]+1，则：dp[index[i]+1]=1-dp[index[i]]