【动态规划中级】AvoidRoads

Problem Statement for AvoidRoads Problem Statement Problem contains images. Plugin users can view them in the applet.[/i] In the city, roads are arranged in a grid pattern. Each point on the grid represents a corner where two blocks meet. The points are connected by line segments which represent the various street blocks. Using the cartesian coordinate system, we can assign a pair of integers to each corner as shown below. You are standing at the corner with coordinates 0,0. Your destination is at corner width[/b],height[/b]. You will return the number of distinct paths that lead to your destination. Each path must use exactly width+height[/b]blocks. In addition, the city has declared certain street blocks untraversable. These blocks may not be a part of any path. You will be given a String[] bad[/b] describing which blocks are bad. If (quotes for clarity) "a b c d" is an element of bad[/b], it means the block from corner a,b to corner c,d is untraversable. For example, let's saywidth[/b] = 6length[/b] = 6bad[/b] = {"0 0 0 1","6 6 5 6"}The picture below shows the grid, with untraversable blocks darkened in black. A sample path has been highlighted in red. Definition Class: AvoidRoads Method: numWays Parameters: int, int, String[] Returns: long Method signature: long numWays(int width, int height, String[] bad) (be sure your method is public) Constraints - width[/b] will be between 1 and 100 inclusive. - height[/b] will be between 1 and 100 inclusive. - bad[/b] will contain between 0 and 50 elements inclusive. - Each element of bad[/b] will contain between 7 and 14 characters inclusive. - Each element of the bad will be in the format "a b c d" where,a,b,c,d are integers with no extra leading zeros, a and c are between 0 and width[/b] inclusive, b and d are between 0 and height[/b] inclusive, and a,b is one block away from c,d. - The return value will be between 0 and 2^63-1 inclusive. Examples 0) 6 6 {"0 0 0 1","6 6 5 6"} Returns: 252 Example from above. 1) 1 1 {} Returns: 2 Four blocks aranged in a square. Only 2 paths allowed. 2) 35 31 {} Returns: 6406484391866534976 Big number. 3) 2 2 {"0 0 1 0", "1 2 2 2", "1 1 2 1"} Returns: 0 来源： <http://community.topcoder.com/stat?c=problem_statement&pm=1889&rd=4709>

收获：1.阻碍是双向的。2.和那个脑筋急转弯的思路一样，路径累加。脑筋急转弯链接见下面附。


从图中可以看出，如果边可达，那么到达roadnum[0][i]=1; roadnum[i][0]=1;到达黄点的路径条数，roadnum[1][1]=roadnum[0][1]+roadnum[1][0]
依次类推，roadnum[i][j]=到达roadnum[i-1][j]和到达road[i][j-1]的路径之和。

#include<stdio.h>
#include<iostream>
using namespace std;
#define LEN 100
int bad[LEN][4];
int badn;
long long roadnum[LEN][LEN];
int reachable(int x1,int y1,int x2,int y2){
for(int i=0;i<badn;i++){
if( (x1==bad[i][0]&& y1==bad[i][1]&& x2==bad[i][2] && y2== bad[i][3]) ||
(x1==bad[i][2]&& y1==bad[i][3]&& x2==bad[i][0]&& y2==bad[i][1]) )
return 0;
}
return 1;
}

long long dp(int m,int n){
for(int i=0;i<=m;i++){
roadnum[i][0]=0;
if(reachable(i,0,i-1,0)&& (i>1&&roadnum[i-1][0]>0 || i==1) )roadnum[i][0]=1;
}
for(int i=0;i<=n;i++){
roadnum[0][i]=0;
if(reachable(0,i,0,i-1)&& (i>1&&roadnum[0][i-1]>0 || i==1) )roadnum[0][i]=1;
}

for(int i=1;i<=m;i++){
for(int j=1;j<=n;j++){
roadnum[i][j]=0;
if(reachable(i,j,i-1,j))
roadnum[i][j]+=roadnum[i-1][j];
if(reachable(i,j,i,j-1))
roadnum[i][j]+=roadnum[i][j-1];
}
}
return roadnum[m]
;
}
int main(){
freopen("in.txt","r",stdin);
int m,n;
while(cin>>m>>n){
cin>>badn;
for(int i=0;i<badn;i++){
for(int j=0;j<4;j++){
cin>>bad[i][j];
}
}
cout<<dp(m,n)<<endl;
}
}

测试用例：6 620 0 0 16 6 5 61 1035 3102 230 0 1 01 2 2 21 1 2 1
结果：252264064843918665349760
------------------------------------------------------------------------------------------------------------网易公开课脑筋急转弯第八课路径计算 http://v.163.com/movie/2011/8/7/C/M95DFI4U4_M95PIUK7C.html第九课三维路径计算 http://v.163.com/movie/2011/8/T/R/M95DFI4U4_M95PJ6FTR.html
这类矩阵的问题要想着先处理边。正如视频中所讲的那样

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拓展多项式求(x+y)n=？如(x+y)4=1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4

可以看出系数就是紫色的对应的部分。拆开相乘就知道譬如x2y2是由6种途径得到的，和图中所展示的一样，到达x2y2有六种途径。很奇妙！
三维见视频。总结：


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来自为知笔记(Wiz)