【hihocoder 1308】搜索二·骑士问题

【题目链接】:http://hihocoder.com/problemset/problem/1308

【题意】

【题解】

用bfs处理出3个骑士到每个点的最短路;
然后枚举最后3个骑士到了哪一个点.
把3个骑士的最短路加起来取最小值就好;

【Number Of WA

0

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,2,2,1,-1,-2,-2,-1,1};
const int dy[9] = {0,-1,1,2,2,1,-1,-2,-2};
const double pi = acos(-1.0);
const int N = 10;

int t;
char s[4][4];
pii a[4];
int dis[4][N][N];
queue <pii> dl;

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    ios::sync_with_stdio(false),cin.tie(0);//scanf,puts,printf not use
    ms(dis,0);
    cin >> t;
    while (t--)
    {
        rep1(i,1,3)
            rep1(j,1,8)
                rep1(k,1,8)
                    dis[i][j][k] = -1;
        rep1(i,1,3)
        {
            cin >> s[i];
            a[i].fi = s[i][0]-'A'+1,a[i].se = s[i][1]-'0';
        }
        rep1(i,1,3)
        {
            dl.push(a[i]);
            dis[i][a[i].fi][a[i].se] = 0;
            while (!dl.empty())
            {
                int x = dl.front().fi,y = dl.front().se;dl.pop();
                rep1(j,1,8)
                {
                    int tx = x+dx[j],ty = y+dy[j];
                    if (dis[i][tx][ty]==-1)
                    {
                        dis[i][tx][ty] = dis[i][x][y]+1;
                        dl.push(mp(tx,ty));
                    }
                }
            }
        }
        int ans = dis[1][1][1]+dis[2][1][1]+dis[3][1][1];
        rep1(i,1,8)
            rep1(j,1,8)
                ans = min(ans,dis[1][i][j]+dis[2][i][j]+dis[3][i][j]);
        cout << ans << endl;
    }
    return 0;
}
posted @ 2017-10-04 18:44  AWCXV  阅读(108)  评论(0编辑  收藏  举报