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HDU 3853 LOOPS (期望DP)

题意:给定一个 n * m的矩阵,然后你从 (1,1)到 (n,m),每次你有三种可能,不动,向右,向下,每次要消耗2个魔法,并且给定每个概率,

问你走出去的期望。

析:dp[i][j] 表示从 (i,j)到终点的概率。然后一路逆推回去就好。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>//#include <tr1/unordered_map>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;//using namespace std :: tr1;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e3 + 5;const LL mod = 10000000000007;const int N = 1e6 + 5;const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){    return r >= 0 && r < n && c >= 0 && c < m;}double dp[maxn][maxn];double a[maxn][maxn][3];int main(){    while(scanf("%d %d", &n, &m) == 2){        for(int i = 0; i < n; ++i)            for(int j = 0; j < m; ++j)                scanf("%lf %lf %lf", &a[i][j][0], &a[i][j][1], &a[i][j][2]);        dp[n-1][m-1] = 0.0;        for(int i = n-1; i >= 0; --i)            for(int j = m-1; j >= 0; --j){                if(i == n-1 && j == m-1)  continue;                if(fabs(1.0 - a[i][j][0]) < eps)  continue;                dp[i][j] = ((2+dp[i][j+1])*a[i][j][1] + (2+dp[i+1][j])*a[i][j][2] + 2*a[i][j][0]) / (1.0-a[i][j][0]);            }        printf("%.3f\n", dp[0][0]);    }    return 0;}

 

HDU 3853 LOOPS (期望DP)