zoukankan      html  css  js  c++  java
  • 【搜索】POJ-3050 基础DFS

    一、题目

    Description

    The cows play the child's game of hopscotch in a non-traditional way. Instead of a linear set of numbered boxes into which to hop, the cows create a 5x5 rectilinear grid of digits parallel to the x and y axes. 
    They then adroitly hop onto any digit in the grid and hop forward, backward, right, or left (never diagonally) to another digit in the grid. They hop again (same rules) to a digit (potentially a digit already visited). 
    With a total of five intra-grid hops, their hops create a six-digit integer (which might have leading zeroes like 000201). 
    Determine the count of the number of distinct integers that can be created in this manner.

    Input

    * Lines 1..5: The grid, five integers per line

    Output

    * Line 1: The number of distinct integers that can be constructed

    Sample Input

    1 1 1 1 1
    1 1 1 1 1
    1 1 1 1 1
    1 1 1 2 1
    1 1 1 1 1
    

    Sample Output

    15
    

    Hint

    OUTPUT DETAILS: 
    111111, 111112, 111121, 111211, 111212, 112111, 112121, 121111, 121112, 121211, 121212, 211111, 211121, 212111, and 212121 can be constructed. No other values are possible.

    二、思路&心得

    • 利用set集合保证数据不重复,最后直接输出set的大小即可
    • 基础深搜,数据量较弱

    三、代码

    #include<cstdio>
    #include<set>
    #define MAX_SIZE 5
    using namespace std;
    
    set<int> s;
    
    int N, M;
    
    int a[10];
    
    int map[MAX_SIZE][MAX_SIZE];
    
    int dirction[4][2] = {0, -1, -1, 0, 0, 1, 1, 0};
    
    bool isLegal(int x, int y) {
    	if (x >= 0 && x < MAX_SIZE && y >= 0 && y < MAX_SIZE) return true;
    	else return false;
    }
    
    void dfs(int x, int y, int k) {
    	if (k == 6) {
    		int num = a[0];
    		for (int j = 1; j < 6; j ++) {
    			num = num * 10 + a[j];
    		}
    		s.insert(num);
    		return;
    	}
    	for(int i = 0; i < 4; i ++) {
    		int tx = x + dirction[i][0], ty = y + dirction[i][1];
    		if (isLegal(tx, ty)) {
    			a[k] = map[tx][ty];
    			dfs(tx, ty, k + 1);
    		}
    	}
    } 
    
    int main() {
    	for (int i = 0; i < 5; i ++) {
    		for (int j = 0; j < 5; j ++) {
    			scanf("%d", &map[i][j]);
    		}
    	}
    	for (int i = 0; i < 5; i ++) {
    		for (int j = 0; j < 5; j ++) {
    			a[0] = map[i][j];
    			dfs(i, j, 1);
    		}
    	}
    	printf("%d", s.size());
    	return 0;
    }
    
  • 相关阅读:
    Tomcat vs Jetty vs Undertow性能对比
    实例对象( instance)、类对象(class)、元类对象(meta-class)的内部结构分析
    isa和superclass
    iOS-weak关键字使用场景
    iOS-weak和assign区别,copy和strong的区别和应用
    iOS-class修饰符的解释及用法
    iOS-atomic修饰符原理剖析讲解 (你将会了解到什么是优先级翻转、自旋锁、互斥锁)
    @property修饰符种类
    @property、@synthesize 、@dynamic的应用
    【原创】Kafka Consumer多线程消费
  • 原文地址:https://www.cnblogs.com/CSLaker/p/7281237.html
Copyright © 2011-2022 走看看