有两个好友A和B，住在一片长有蘑菇的由n＊m个方格组成的草地，A在(1,1),B在(n,m)。现在A想要拜访B，由于她只想去B的家，所以每次她只会走(i,j+1)或(i+1,j)这样的路线，在草地上有k个蘑菇种在格子里(多个蘑菇可能在同一方格),问：A如果每一步随机选择的话(若她在边界上，则只有一种选择)，那么她不碰到蘑菇走到B的家的概率是多少？

第二种方法：首先分析题意，可用概率的方法来计算，做了好几道百度的题目，觉得大多数是再考概率论，所以首先要弄懂题意，最后做题前把公式写出来，这样编码时才能游刃有余。

本题中下面的第一种用迭代枚举的方法来做是不对的，仅做错误示范

// ConsoleApplication3.cpp : 定义控制台应用程序的入口点。
//

#include<iostream>
#include<vector>
#include<iomanip>
using namespace std;

int main()
{
	int M, N, K;

	while (cin >> N >> M >> K)
	{
		vector<vector<int>> lawn(N, vector<int>(M, 0));
		vector<vector<float>> pVec(N, vector<float>(M, 0));
		pVec[0][0] = 1;
		for (int i = 0; i < K; i++)
		{
			int x, y;
			cin >> x >> y;
			lawn[x - 1][y - 1] = 1;
		}
		pVec[0][0] = 1;
		for (int x = 0; x < N; x++)
		{
			for (int y = 0; y < M; y++)
			{
				if (x == 0 && y == 0)
				{
					//		cout << "x:" << x << "y：" << y << " " << "初始化" << endl;
					pVec[0][0] = 1.0;

				}
			   else if (lawn[x][y] == 1)//中间有障碍物
				{
				//	cout << "x:" << x << "y：" << y << "有障碍 "  << endl;
					pVec[x][y] = 0;
				}

				else if (x < lawn.size() - 1 && y < lawn[0].size() - 1) //中间
				{
				//	cout <<"x:"<<x<<"  y："<<y<<" " <<"中间" << endl;

					if (x - 1 >= 0)
					{
				//		cout << "#####" << endl;
						pVec[x][y] = pVec[x][y] + pVec[x - 1][y] * 0.5;
				//		cout << "x-1:" << x-1 << "  y：" << y << " " << pVec[x-1][y] << endl;
				//		cout << "x:" << x << "  y：" << y << " " << pVec[x][y] << endl;
					}
					if (y - 1 >= 0)
					{
						pVec[x][y] = pVec[x][y] + pVec[x][y - 1] * 0.5;
					}
				}
				else if (x < lawn.size() - 1 && y == lawn[0].size() - 1) //最右面
				{
		//			cout << "x:" << x << "y：" << y << " " << "最右面" << endl;
					if (x - 1 >= 0)
					{
						pVec[x][y] = pVec[x][y] + pVec[x - 1][y] * 1;
					}
					if (y - 1 >= 0)
					{
						pVec[x][y] = pVec[x][y] + pVec[x][y - 1] * 0.5;
					}
				}
				else if (x == lawn.size() - 1 && y < lawn[0].size() - 1) //最下面
				{
			//		cout << "x:" << x << "y：" << y << " " << "最下面" << endl;
					if (x - 1 >= 0)
					{
						pVec[x][y] = pVec[x][y] + pVec[x - 1][y] * 0.5;
					}
					if (y - 1 >= 0)
					{
						pVec[x][y] = pVec[x][y] + pVec[x][y - 1] * 1;
					}
				}
				else  //右下角
				{
		//			cout << "x:" << x << "y：" << y << " " << "右下角" << endl;
					if (x - 1 >= 0)
					{
						pVec[x][y] = pVec[x][y] + pVec[x - 1][y] * 1;

					}
					if (y - 1 >= 0)
					{
						pVec[x][y] = pVec[x][y] + pVec[x][y - 1] * 1;
						//cout << "下面" << endl;
					}
				}

			}

		}
		/*for (int i = 0; i < pVec.size(); i++)
		{
			for (int j = 0; j < pVec[0].size(); j++)
			{
				cout << pVec[i][j] << "  ";
			}
		}*/
	//	cout << endl;
		cout << setiosflags(ios::fixed);
		cout << setprecision(2)<<pVec[N-1][M-1] << endl;

	}
	return 0;
}

第一种方法：迭代，枚举每条路径，但是当n或m大于10时算法就很慢了，这种方法不适用。当数大于等于9时，考虑使用第二种方法，和这个题目类似的是，京东出的题，小东发年终奖，但是那个题目是在6*6的棋盘上，不用担心算法复杂度，对这个题目不适合。。。。。。。

#include "stdafx.h"
#include<iostream>
#include<vector>
#include<iomanip>
using namespace std;
class Solution {
public:
	//没有障碍

	int noObstacle(vector<vector<int>> lawn)
	{

		if (lawn.size() == 1)
		{

			return 1;
		}
		else if (lawn[0].size() == 1)
		{

			return 1;
		}
		else
		{

			return  deleteRow(lawn) + deleteCol(lawn);
		}

	}

	int  deleteRow(vector<vector<int>> lawn)
	{
		lawn.erase(lawn.begin());
		return	 noObstacle(lawn);
	}

	int  deleteCol(vector<vector<int>> lawn)
	{

		for (int i = 0; i < lawn.size(); i++)
		{
			lawn[i].erase(lawn[i].begin());
		}
		return	noObstacle(lawn);
	}

	//有障碍

	int hasObstacle(vector<vector<int>> lawn)
	{
		if (lawn.size() == 1)
		{
			bool flag = false;
			for (int i = 0; i < lawn[0].size(); i++)
			{
				if (lawn[0][i] != 0)
				{
					flag = true;
					break;
				}
			}
			if (flag == false) return 0;
			return 1;
		}
		else if (lawn[0].size() == 1)
		{
			bool flag = false;
			for (int i = 0; i < lawn.size(); i++)
			{
				if (lawn[i][0] != 0)
				{
					flag = true;
					break;
				}
			}
			if (flag == false) return 0;
			return 1;
		}
		else
		{

			return  deleteRow2(lawn) + deleteCol2(lawn);
		}
	}
	int  deleteRow2(vector<vector<int>> lawn)
	{
		if (lawn[1][0] != 0)
		{
			lawn.erase(lawn.begin());
			return	 noObstacle(lawn);
			return 0;
		}
		else
		{
			lawn.erase(lawn.begin());
			return	 hasObstacle(lawn);
		}

	}

	int  deleteCol2(vector<vector<int>> lawn)
	{
		if (lawn[0][1] != 0)
		{
			for (int i = 0; i < lawn.size(); i++)
			{
				lawn[i].erase(lawn[i].begin());
			}
			return	 noObstacle(lawn);
			return 0;
		}
		else
		{
			for (int i = 0; i < lawn.size(); i++)
			{
				lawn[i].erase(lawn[i].begin());
			}
			return	 hasObstacle(lawn);
		}

	}

};

int main()
{
	int M, N, K;
	Solution so;
	while (cin>>N>>M>>K)
	{
		vector<vector<int>> lawn;
		for (int i = 0; i < N; i++)
		{
			vector<int> vec;
			for (int j = 0; j < M; j++)
			{
				vec.push_back(0);
			}
			lawn.push_back(vec);
		}

		for (int i = 0; i < K; i++)
		{
			int x,y;
			cin >> x >> y;
			lawn[x - 1][y - 1]= lawn[x - 1][y - 1]+1;
		}

		cout << setiosflags(ios::fixed);
		cout << "有障碍：" << so.hasObstacle(lawn)<<endl;
		cout << "无障碍：" << so.noObstacle(lawn)  << endl;

		float p = 1- ((float)so.hasObstacle(lawn)) / ((float)so.noObstacle(lawn));

		cout << setprecision(2)<< p << endl;
	}

	return 0;
}