Rokua OJ P1141 01 Maze Problem Solving Report

Rokua OJ P1141 01 Maze Problem Solving Report

by Medalplus

"title description"

There is a labyrinth of nxn that consists only of numbers 0 and 1. If you are on a grid of 0, then you can move to one of the next 4 grid 1, also if you are on a grid of 1, then you can move to a 4 grid adjacent to a certain grid 0.
Your task is: For a given maze, ask how many squares (including itself) you can move from one grid to the beginning.

"Input description"

The 1th behavior of the input is two positive integer n,m.
The following n lines, n characters per line, can only be 0 or 1 characters, with no spaces between characters.
The next m line, each line of 2 positive integers separated by a space, corresponds to the i,j in the maze of row J of a lattice, asking how many squares can be moved from the beginning of the lattice.

"Output description"

The output includes m lines, and the corresponding answer is output for each query.

Analysis

This question compares the pit father

The problem is to ask for the number of Unicom blocks where a point resides .

Then I think of the Tarjan and the collection and the

The result looked under the range n<=1000,m<=100000, then gave up ...

Think about it, it's a BFS.

Then find that the time complexity is O (n2m)

And then the tle, we have to consider optimizing the

We find that the number of squares that can go on the path of a point is the same !

So that means we can preprocess the entire graph.

It reminds us of Flood_fill.

Then the author uses the DFS approach

Then the stack, found that the full picture of the 1000*1000 (all 1 or 0) will reach 106 layers, so it will explode

This time may consider the BFS way Flood_fill

Then on the tle, I found the main slow time is to take the STL's queue

Special time consuming, and then write a queue, and then add the card constant read into

It's AC ....

A hint of sadness

"Code"

1#include <iostream>2#include <cstdio>3#include <cstring>4 using namespacestd;5 6 #defineRep (I,l,r) for (i=l;i<=r;i++)7 #defineRead scanf8 #definePrint printf9 Ten Const intmaxn=1001; One Const intmaxm=10000001; A  - structxoy{ -     intx, y; the }; -  - CharGRAPH[MAXN][MAXN]; - intRECORD[MAXM],ID[MAXN][MAXN]; + xOy Q[MAXM]; - intFront,tail; + intN,m;//size of the board; question numbers A intDatax,datay; at  - voidBFsintXintYintindex) { -Front=tail=1; -Tail=2; -q[1]=(structxOy) {x, y}; - xOy head; in      while(front!=tail) { -Head=Q[front]; to++Front; +         if(Id[head.x][head.y])Continue; -id[head.x][head.y]=index; therecord[index]++; *         if(head.x-1>=1&& graph[head.x-1][head.y]!=Graph[head.x][head.y]) $Q[tail++]= (structXOy) {head.x-1, head.y};Panax Notoginseng         if(head.y-1>=1&& graph[head.x][head.y-1]!=Graph[head.x][head.y]) -Q[tail++]= (structXOy) {head.x,head.y-1}; the         if(head.x+1<=n && graph[head.x+1][head.y]!=Graph[head.x][head.y]) +Q[tail++]= (structXOy) {head.x+1, head.y}; A         if(head.y+1<=n && graph[head.x][head.y+1]!=Graph[head.x][head.y]) theQ[tail++]= (structXOy) {head.x,head.y+1}; +     }  - } $  $ intMain () { -Read"%d%d",&n,&m); -     inti,j,index=1; theRep (I,1, N) -Rep (J,1, N)WuyiCin>>Graph[i][j]; theRep (I,1, N) -Rep (J,1, N) Wu       { - BFS (i,j,index); About         if(Record[index]) index++; $       } -Rep (I,1, M) { -Read"%d%d",&datax,&Datay); -Print"%d\n", Record[id[datax][datay]]); A     } +     return 0; the}

Rokua OJ P1141 01 Maze Problem Solving Report