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 ..*..
.......
.......
...... 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);
}