HDU 1026 search pruning problem within the labyrinth of Ignatius and the prince I

Hello everyone , I meet you again , I'm the king of the whole stack .

This problem is a typical type of problem maze extensive Search for .

I see a lot of solutions on the Internet .

There's no problem-solving analysis , Don't point out the key points . The code is more like a copy . Some analysts also have big articles to analyze . Just don't hit the key all the time , What is broad and profound , Even search and find , Analyzing differences . Why search wide image quickly , What kind of storm calls search , All wrong. . The code is copied .

Through a long period of research , It finally got through .

Although this topic can be said to be extensive search . But the key is pruning . Why? ？

Because the maze can't search all the paths simply by searching widely , Even if the maze is a little bigger, you can't find out whether there is a path . Suppose there is no conditional pruning . Don't believe it , You strictly write a wide search and search the maze path to see . Of course, you wrote a wrong search . Naturally, I got the wrong answer .

The common mistake is to expand the maze one by one, thinking it is a wide search of the maze , wrong ！

The real search is to map the maze . Then search widely .

In fact, the real key is pruning ：

Pruning thinking is to think about when it should be extended to the next grid ？ Whether the lattice is legal or not must be able to expand ？ Of course not. , It is necessary to prune according to the meaning of the topic . The meaning of this question is to find the path with the smallest time . Therefore, it can be thought that it should be decided by time comparison whether it needs to be extended to the next grid .

That is, if the next lattice is possible to find a better solution, it will expand . Otherwise, do not expand .

After such a pruning . You can use the so-called wide search .

So why this topic . Or it can be said that the topic of this topic type cannot use deep search ？

Because the above pruning condition is that each layer is pruned , Suppose you use deep search , Then this pruning condition cannot be used .

Another way is to use priority queues . Give priority to expanding the lattice with the current minimum time . Then you don't have to search the next box repeatedly . This is also the use of the above pruning ideas .

Just understand the key pruning points above , Then this kind of problem can be solved at will .

As for the recording path of this question, it is the test of programming skills , Needless to say, what ideas . No, just say that the coding ability is not good .

Maybe some people who don't understand analysis also hit the code right , Maybe he's lucky . Maybe he is really a genius ！

Anyway, the probability is very low , The greatest chance is that his code is copied .

#include <stdio.h>
#include <vector>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <string>
#include <limits.h>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;

/*
 Key understanding ： Only when the next grid updates the minimum value does it need to be extended to this grid . Otherwise, there is no need to expand to this lattice . This is also equivalent to the pruning point of guangsearch . I can't understand this . There is no thorough understanding of this problem .*/namespace IgnatiusandthePrincessI1026{const int MAX_N = 101;char Maze[MAX_N][MAX_N];int dx[] = {-1, 0, 1, 0}, dy[] = {0, -1, 0, 1};struct Node{	int sec, x, y;	Node *p;};Node mazeRec[MAX_N][MAX_N];int N, M;inline bool isLegal(int r, int c){	return 0<=r && 0<=c && r<N && c<M && Maze[r][c] != 'X';}inline int getSec(int r, int c){	if (Maze[r][c] == '.') return 1;	return Maze[r][c] - '0' + 1;}void getPath(){	for (int i = 0; i < N; i++)	{		for (int j = 0; j < M; j++)		{			mazeRec[i][j].sec = INT_MAX;			mazeRec[i][j].x = i, mazeRec[i][j].y = j;			mazeRec[i][j].p = NULL;		}	}	queue<Node *> qu;	Node *p = &mazeRec[N-1][M-1];	// Pay attention to the calculation error ：p->sec = Maze[N-1][M-1] or = getSec(N-1, M-1)	p->sec = getSec(N-1, M-1)-1;// The end may also be an enemy , The starting point stipulates invincible 	qu.push(p);	while (!qu.empty())	{		p = qu.front(); qu.pop();		for (int i = 0; i < 4; i++)		{			int tx = p->x + dx[i], ty = p->y + dy[i];			if (!isLegal(tx, ty)) continue;			int sec = getSec(tx, ty);			Node *tmp = &mazeRec[tx][ty];			if (p->sec+sec < tmp->sec)			{				tmp->sec = p->sec+sec;				tmp->p = p;				qu.push(tmp);			}			/* Key understanding ： Only when the next grid updates the minimum value does it need to be extended to this grid . Otherwise, there is no need to expand to this lattice . This is also equivalent to the pruning point of guangsearch . I can't understand this , There is no thorough understanding of this problem .*//* All kinds of mistakes and lessons ！			qu.push(tmp);			tmp.vis = true;	// Multiple errors else. Logic error else tmp->vis = true			//Maze[tx][ty] = 'X';			tmp.sec = p.sec+sec;			tmp.p = &mazeRec[p.x][p.y];						// error ：tmp->p = p;			// error ：tmp->sec = min(tmp->sec, p->sec+sec);*/		}	}}int main(){	while (~scanf("%d %d", &N, &M))	{		while (getchar() != '\n');		for (int i = 0; i < N; i++)		{			gets(Maze[i]);		}		getPath();		Node *p = &mazeRec[0][0];		if (p->sec == INT_MAX) puts("God please help our poor hero.");		else		{			printf("It takes %d seconds to reach the target position, let me show you the way.\n", p->sec);			int s = 1;			for (; p->p; p = p->p)			{				int x = p->p->x, y = p->p->y;				printf("%ds:(%d,%d)->(%d,%d)\n", s++, p->x, p->y, x, y);				if (Maze[x][y] == '.') continue;				int fig = Maze[x][y] - '0';// Fewer mistakes -'0'				for (int i = 0; i < fig; i++)				{					printf("%ds:FIGHT AT (%d,%d)\n", s++, x, y);				}			}		}		puts("FINISH");	}	return 0;}

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