xutian_curry的博客

思路是广度优先搜索，每次记录路径和距离#include<stdio.h>#include<stdlib.h>#include <string.h>#include <string>#include <iostream>#include <stack>#include <math.h>#include

广搜，值得学习的地方时是用二进制数来存储已获得的钻石种类。#include <iostream>#include <memory.h>#include <vector>#include <queue>#include <stdio.h>#include <algorithm>#include <unorder...

题目大意：巨大流星雨即将袭来。每个流星会对击中的地方以及周围（上下左右四格）造成破坏。Bessie开始时位于（0, 0）位置，并希望逃到一处不会被袭击到的地方（在第一象限内）。已知每移动一格需要1个时间单位，被流星破坏后的地方不能再进入。给出M个流星在T时刻击中的地方（X, Y），问Bessie能否逃到安全的地方，若能输出最短时间，否则输出-1。裸的广搜，注意判重即可#include <io...

用优先级队列来存放准备扩展的结点，时间少的在队首#include <iostream>#include <memory.h>#include <string>#include <stdio.h>#include <queue>#include <math.h>#include <stdlib.h>us

#include <iostream>#include <memory.h>#include <string>#include <stdio.h>#include <queue>#include <math.h>#include <stdlib.h>using namespace std;int R

当队列为空时，则表示到达不了指定位置#include <iostream>#include <memory.h>#include <string>#include <stdio.h>#include <queue>#include <math.h>#include <stdlib.h>using na

#include <iostream>#include <memory.h>#include <string>#include <stdio.h>#include <queue>using namespace std;char Map[45][45];int r,c;int visited[45][45];int ste...

单源无权最短路径，利用广度优先搜索#include<stdio.h>#include<stdlib.h>#include <string.h>#include <string>#include <iostream>#include <stack>#include <math.h>#include &a

类似于层序遍历，用一个队列非递归实现#include <iostream>#include <vector>#include <stdio.h>#include <memory.h>#include <queue>using namespace std;int visited[9];vector < vector &l...

#include<iostream>#include <stdio.h>#include <memory.h>#include <queue>using namespace std;int M,N,T;char Map[205][205];int visited[205][205][15];int dx[]={1,-1,0,0};i...

#include<iostream>using namespace std;struct node{ int x,y,pre;};node a[100];int m[6][6];int dx[]= {1,-1,0,0};int dy[]= {0,0,1,-1};int head,tail;void print(int i){ if(a[i].pre!...

裸的广搜，visited数组可以用一个map代替#include <iostream>#include <string>#include <string.h>#include <stdio.h>#include <malloc.h>#include <math.h>#include <map>#inc