广搜 广搜 poj 3984

***求最短路径，然后再输出路径，

BFS+路径输出***

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <cmath>using namespace std;
typedef long long LL;
#define oo 0x3f3f3f3f
#define N 100int maps[10][10];
int dir[4][2]= {{1,0}, {0,1}, {-1,0}, {0,-1}}, vis[N][N], step[N][N];struct node
{int x, y, s;} a, b;int k;void BFS(int x, int y)
{k=0;queue<node> q;a.x=x;a.y=y;a.s=0;q.push(a);vis[0][0]=1;step[0][0]=0;k++;while(!q.empty()){a=q.front();q.pop();for(int i=0; i<4; i++){b.x=a.x+dir[i][0];b.y=a.y+dir[i][1];if(b.x>=0&&b.x<5&&b.y>=0&&b.y<5&&maps[b.x][b.y]==0&&!vis[b.x][b.y]){b.s=a.s+1;q.push(b);step[b.x][b.y]=b.s;vis[b.x][b.y]=1;}
        }}
}void put(int x, int y)
{if(x==0 && y==0){printf("(%d, %d)\n", x, y);return ;}for(int i=0; i<4; i++){int nx=x+dir[i][0];int ny=y+dir[i][1];if(nx>=0 && nx<5 && ny>=0 && ny<5 && step[x][y]==step[nx][ny]+1 && vis[nx][ny]){put(nx, ny);printf("(%d, %d)\n", x, y);}}
}int main()
{for(int i=0; i<5; i++)for(int j=0; j<5; j++)scanf("%d", &maps[i][j]);memset(vis, 0, sizeof(vis));memset(step, 0, sizeof(step));BFS(0, 0);put(4, 4);return 0;
}