POJ3744（概率dp）

思路：一长段概率乘过去最后会趋于平稳，所以因为地雷只有10个，可以疯狂压缩其位置，这样就不需要矩阵乘优化了。另外初始化f[0] = 0, f[1] = 1，相当于从1开始走吧。双倍经验：洛谷1052.

 for (int i = ; i <= n; i++) {
     if (pos[i] - pos[i - ] > ) {
         for (int j = n; j >= i; j--)
             pos[j] -= (pos[i] - pos[i - ] - );
     }
 }

这段代码j要倒着写否则先从i开始的话pos[i] - pos[i-1]就变了，我tm居然WA了一板一上午……请叫我绝世大傻逼。

非矩阵版：

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

 typedef double db;

 const int maxn = 1e3 + ;
 int n, pos[];
 db f[maxn];
 db p;

 db solve() {
     memset(f, , sizeof f);
     f[] = 1.0;
     int t = ;
     for (int i = ; i <= pos[n]; i++) {
         if (i == pos[t])    f[i] = , t++;
         f[i + ] += p * f[i];
         f[i + ] += (1.0 - p) * f[i];
     }
     return f[pos[n] + ];
 }

 int main() {
     while (scanf("%d%lf", &n, &p) != EOF) {
         for (int i = ; i <= n; i++)    scanf("%d", &pos[i]);
         sort(pos + , pos +  + n);
         for (int i = ; i <= n; i++) {
             if (pos[i] - pos[i - ] > ) {
                 for (int j = n; j >= i; j--)
                     pos[j] -= (pos[i] - pos[i - ] - );
             }
         }
         printf("%.7f\n", solve());
     }
     return ;
 }

顺手练一下矩阵，写得low了点：

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <ctime>
 #include <cctype>
 #include <climits>
 #include <iostream>
 #include <iomanip>
 #include <algorithm>
 #include <string>
 #include <sstream>
 #include <stack>
 #include <queue>
 #include <set>
 #include <map>
 #include <vector>
 #include <list>
 #include <fstream>
 #include <bitset>
 #define init(a, b) memset(a, b, sizeof(a))
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define irep(i, a, b) for (int i = a; i >= b; i--)
 using namespace std;

 typedef double db;
 typedef long long ll;
 typedef unsigned long long ull;
 typedef pair<int, int> P;
 const int inf = 0x3f3f3f3f;
 const ll INF = 1e18;

 template <typename T> void read(T &x) {
     x = ;
     int s = , c = getchar();
     for (; !isdigit(c); c = getchar())
         if (c == '-')    s = -;
     for (; isdigit(c); c = getchar())
         x = x *  + c - ;
     x *= s;
 }

 template <typename T> void write(T x) {
     if (x < )    x = -x, putchar('-');
     if (x > )    write(x / );
     putchar(x %  + '');
 }

 template <typename T> void writeln(T x) {
     write(x);
     puts("");
 }

 struct Matrix {
     db v[][];

     Matrix() {
         init(v, );
     }

     friend Matrix operator * (Matrix A, Matrix B) {
         Matrix ret;
         rep(i, , )    rep(j, , )    rep(k, , )
             ret.v[i][j] += A.v[i][k] * B.v[k][j];
         return ret;
     }

     friend Matrix operator ^ (Matrix A, int k) {
         Matrix ret;
         ret.v[][] = ret.v[][] = ;
         for (; k; k >>= ) {
             if (k & )    ret = ret * A;
             A = A * A;
         }
         return ret;
     }
 };

 int n, pos[];
 db p;

 int main() {
     while (~scanf("%d%lf", &n, &p)) {
         rep(i, , n)    read(pos[i]);
         sort(pos + , pos +  + n);

         pos[] = , pos[n + ] = pos[n] + ;
         Matrix f, dp;
         f.v[][] = ;
         rep(i, , n + ) {
             int t = pos[i] - pos[i - ];
             dp.v[][] = p, dp.v[][] =  - p, dp.v[][] = , dp.v[][] = ;
             f = f * (dp ^ t);
             if (i <= n)    f.v[][] = ;
         }

         printf("%.7f\n", f.v[][]);
     }
     return ;
 }