CSU 1511 残缺的棋盘

Problem I: 残缺的棋盘

Time Limit: 1 Sec Memory Limit:
128 MB

Submit: 526 Solved: 186

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Description

Input

输入包含不超过10000 组数据。每组数据包含6个整数r1, c1, r2, c2, r3, c3 (1<=r1, c1, r2, c2, r3, c3<=8). 三个格子A, B, C保证各不相同。

Output

对于每组数据，输出测试点编号和最少步数。

Sample Input

1 1 8 7 5 6
1 1 3 3 2 2

Sample Output

Case 1: 7
Case 2: 3

简单BFS搜索。判断条件里加上r3,c3就行了。

#include <stdio.h>
#include <queue>
#include <string.h>
#define N 8
using namespace std;
typedef struct node{
int position, step;
}node;
queue<node> q;
int r1, c1, r2, c2, r3, c3;
int dx[8] = { 0, 1, 1, -1, -1, 0, -1, 1 };
int dy[8] = { 1, 1, 0, 1, 0, -1, -1, -1 };
int vis

;
int kstart,kend;
int bfs(int pos)
{
while (!q.empty())
q.pop();
node now, next;
now.position = pos;
now.step = 0;
vis[now.position / N][now.position%N] = 1;
q.push(now);
while (!q.empty())
{
node front = q.front();
q.pop();
if (((front.position) / N) == (kend/N) && ((front.position) % N) == kend%N)
{
return front.step;
}
int x = front.position / N;
int y = front.position%N;
for (int i = 0; i < 8; i++)
{
int xx = x + dx[i];
int yy = y + dy[i];
next.position = xx*N + yy;
if (xx < 0 || yy < 0 || xx >= N || yy >= N || vis[xx][yy] == 1||(xx==r3&&yy==c3))
continue;
vis[xx][yy] = 1;
next.step = front.step + 1;
q.push(next);
}
}
return -1;
}
int main()
{
int i=1;
while(scanf("%d%d%d%d%d%d",&r1,&c1,&r2,&c2,&r3,&c3)>0)
{
memset(vis,0,sizeof(vis));
r1--,c1--,r2--,c2--,r3--,c3--;
kstart=r1*N+c1;
kend=r2*N+c2;
int ans=bfs(kstart);
printf("Case %d: %d\n",i++,ans);
}
return 0;
}

/**************************************************************
Problem: 1511
User: lizhaowei213
Language: C++
Result: Accepted
Time:120 ms
Memory:1064 kb
****************************************************************/