bzoj3393 [Usaco2009 Jan]Laserphones 激光通讯

Description

Input

第1行输入w和H，之后W行H列输入地图，图上符号意义如题目描述．

Output

最少的对角镜数量．

Sample Input

7 8
.......
...... C
......*
*****.*
....*..
....*..
.C ..*..
.......

Sample Output

3

求拐点数最小

直接搜索是不行的，因为可能出现一个点当前被更新的状态并不是它最优的状态

所以像spfa那样允许多次入队，这样虽然慢一点但是没有后效性，而且这么小的数据也不会慢到哪里去吧

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#define LL long long
#define inf 598460606
#define pa pair<int,int>
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
struct que{int x,y,dire,dist;}now,wrk;
bool operator < (const que &a,const que &b)
{return a.dist>b.dist;}
priority_queue <que> q;
const int mx[4]={0,1,0,-1};
const int my[4]={1,0,-1,0};
int n,m,sx,sy,ex,ey;
bool mrk[110][110];
int dist[110][110][4];
int main()
{
    scanf("%d%d",&m,&n);
    for (int i=1;i<=n;i++)
      for (int j=1;j<=m;j++)
      {
        char ch=getchar();
        while (ch!=‘C‘&&ch!=‘.‘&&ch!=‘*‘)ch=getchar();
        if (ch==‘*‘)mrk[i][j]=1;
        if (ch==‘C‘){if (!sx){sx=i;sy=j;}else {ex=i;ey=j;} }
      }
    memset(dist,127,sizeof(dist));
    now.x=sx;now.y=sy;now.dist=0;
    for (int i=0;i<4;i++)
    {
        now.dire=i;
        q.push(now);
        dist[sx][sy][i]=0;
    }
    while (!q.empty())
    {
        now=q.top();q.pop();
        int k=now.dire;
        wrk=now;
        while (wrk.x+mx[k]>=1&&wrk.x+mx[k]<=n&&wrk.y+my[k]>=1&&wrk.y+my[k]<=m&&!mrk[wrk.x+mx[k]][wrk.y+my[k]]&&dist[wrk.x+mx[k]][wrk.y+my[k]][k]>wrk.dist)
        {
            wrk.x+=mx[k];wrk.y+=my[k];
            dist[wrk.x][wrk.y][k]=dist[now.x][now.y][k];
            q.push(wrk);
        }
        wrk=now;wrk.dist++;
        for (int k=0;k<4;k++)
        if(dist[now.x][now.y][k]>now.dist+1)
        {
            dist[now.x][now.y][k]=now.dist+1;
            wrk.dire=k;
            q.push(wrk);
        }
    }
    int ans=inf;
    for (int k=0;k<4;k++)
      ans=min(ans,dist[ex][ey][k]);
    printf("%d\n",ans);
}