2015
04-13

# LOOPS

Akemi Homura is a Mahou Shoujo (Puella Magi/Magical Girl).

Homura wants to help her friend Madoka save the world. But because of the plot of the Boss Incubator, she is trapped in a labyrinth called LOOPS.

The planform of the LOOPS is a rectangle of R*C grids. There is a portal in each grid except the exit grid. It costs Homura 2 magic power to use a portal once. The portal in a grid G(r, c) will send Homura to the grid below G (grid(r+1, c)), the grid on the right of G (grid(r, c+1)), or even G itself at respective probability (How evil the Boss Incubator is)!
At the beginning Homura is in the top left corner of the LOOPS ((1, 1)), and the exit of the labyrinth is in the bottom right corner ((R, C)). Given the probability of transmissions of each portal, your task is help poor Homura calculate the EXPECT magic power she need to escape from the LOOPS.

The first line contains two integers R and C (2 <= R, C <= 1000).

The following R lines, each contains C*3 real numbers, at 2 decimal places. Every three numbers make a group. The first, second and third number of the cth group of line r represent the probability of transportation to grid (r, c), grid (r, c+1), grid (r+1, c) of the portal in grid (r, c) respectively. Two groups of numbers are separated by 4 spaces.

It is ensured that the sum of three numbers in each group is 1, and the second numbers of the rightmost groups are 0 (as there are no grids on the right of them) while the third numbers of the downmost groups are 0 (as there are no grids below them).

You may ignore the last three numbers of the input data. They are printed just for looking neat.

The answer is ensured no greater than 1000000.

Terminal at EOF

The first line contains two integers R and C (2 <= R, C <= 1000).

The following R lines, each contains C*3 real numbers, at 2 decimal places. Every three numbers make a group. The first, second and third number of the cth group of line r represent the probability of transportation to grid (r, c), grid (r, c+1), grid (r+1, c) of the portal in grid (r, c) respectively. Two groups of numbers are separated by 4 spaces.

It is ensured that the sum of three numbers in each group is 1, and the second numbers of the rightmost groups are 0 (as there are no grids on the right of them) while the third numbers of the downmost groups are 0 (as there are no grids below them).

You may ignore the last three numbers of the input data. They are printed just for looking neat.

The answer is ensured no greater than 1000000.

Terminal at EOF

2 2
0.00 0.50 0.50    0.50 0.00 0.50
0.50 0.50 0.00    1.00 0.00 0.00

6.000

## 题目大意

有一个人被困在一个 R*C(2<=R,C<=1000) 的迷宫中，起初他在 (1,1) 这个点，迷宫的出口是 (R,C)。在迷宫的每一个格子中，他能花费 2 个魔法值开启传送通道。假设他在 (x,y) 这个格子中，开启传送通道之后，有 p_lift[i][j] 的概率被送到 (x,y+1)，有 p_down[i][j] 的概率被送到 (x+1,y)，有 p_loop[i][j]

## 做法分析

令：f[i][j] 表示从 (i,j) 这个点到出口 (R,C) 花费的魔法值的期望。

那么，我们有：

f[i][j] = p_loop[i][j]*f[i][j] + p_left[i][j]*f[i][j+1] + p_down[i][j]*f[i+1][j]

移项可得：
(1-p_loop[i][j])*f[i][j] = p_left[i][j](f[i][j+1] + p_down[i][j]*f[i+1][j]

于是我们可以倒着递推了

## 参考代码

#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>

using namespace std;

double p[3][1005][1005];
double f[1005][1005];
bool vs[1005][1005];
int R, C;

double DP(int x, int y)
{
if(x==R && y==C) return 0;
if(x>R || y>C) return 0;
if(vs[x][y]) return f[x][y];
if(abs(p[0][x][y]-1)<1e-6)
{
vs[x][y]=1;
return f[x][y]=0;
}
vs[x][y]=1;
return f[x][y]=(2+p[1][x][y]*DP(x, y+1)+p[2][x][y]*DP(x+1, y))/(1-p[0][x][y]);
}

int main()
{
while(scanf("%d%d", &R, &C)!=EOF)
{
for(int i=1; i<=R; i++)
for(int j=1; j<=C; j++)
scanf("%lf%lf%lf", &p[0][i][j], &p[1][i][j], &p[2][i][j]);
memset(f, 0, sizeof(f));
memset(vs, 0, sizeof(vs));
printf("%.3lf\n", DP(1, 1));
}
return 0;
}

## AC通道

HDU 3853 LOOPS

1. I like your publish. It is great to see you verbalize from the coronary heart and clarity on this essential subject matter can be easily noticed.

2. 代码不对，仔细对比一下输入输出， 怎么会有‘‘printf("The winning moves are:");’’这种输出呢？

3. 换句话说，A[k/2-1]不可能大于两数组合并之后的第k小值，所以我们可以将其抛弃。
应该是，不可能小于合并后的第K小值吧