USACO Laserphones
Description
The cows have a new laser-based system so they can have casual
conversations while out in the pasture which is modeled as a W x H
grid of points (1 <= W <= 100; 1 <= H <= 100).
The system requires a sort of line-of-sight connectivity in order
to sustain communication. The pasture, of course, has rocks and
trees that disrupt the communication but the cows have purchased
diagonal mirrors ('/' and '' below) that deflect the laser beam
through a 90 degree turn. Below is a map that illustrates the
problem.
H is 8 and W is 7 for this map. The two communicating cows are
notated as 'C's; rocks and other blocking elements are notated as
'*'s:
7 . . . . . . . 7 . . . . . . .
6 . . . . . . C 6 . . . . . /-C
5 . . . . . . * 5 . . . . . | *
4 * * * * * . * 4 * * * * * | *
3 . . . . * . . 3 . . . . * | .
2 . . . . * . . 2 . . . . * | .
1 . C . . * . . 1 . C . . * | .
0 . . . . . . . 0 . -------/ .
0 1 2 3 4 5 6 0 1 2 3 4 5 6
Determine the minimum number of mirrors M that must be installed
to maintain laser communication between the two cows, a feat which
is always possible in the given test data.
Input
* Line 1: Two space separated integers: W and H
* Lines 2..H+1: The entire pasture.
Output
* Line 1: A single integer: M
Sample Input
7 8 ....... ......C ......* ****. ...... ...... .C..*.. .......
Sample Output
3
最优解声明：
题解：
思路：
最优化——DP，走地图——搜索。DP+搜索：记忆化搜索。
然后就可以开始码了。
本题的坑点：m,n是反着读的。
代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,ans=10000;
char mp[110][110];
bool v[110][110];
int dp[110][110][5];
int dx[]={0,1,-1,0,0};
int dy[]={0,0,0,-1,1};
void dfs(int x,int y,int now,int d)
{
if(d&&dp[x][y][d]<=now)
return;
if(d)
dp[x][y][d]=now;
if(mp[x][y]=='C')
{
ans=min(ans,now);
return;
}
if(d)
{
int xx=x+dx[d];
int yy=y+dy[d];
if(xx>=1&&yy>=1&&xx<=n&&yy<=m&&!v[xx][yy]&&mp[xx][yy]!='*')
{
v[xx][yy]=1;
dfs(xx,yy,now,d);
v[xx][yy]=0;
}
}
for(int i=1;i<=4;i++)
{
int xx=x+dx[i];
int yy=y+dy[i];
if(xx<1||yy<1||xx>n||yy>m||v[xx][yy]||mp[xx][yy]=='*')
continue;
v[xx][yy]=1;
if(!d)
dfs(xx,yy,now,i);
else
dfs(xx,yy,now+1,i);
v[xx][yy]=0;
}
}
int main()
{
scanf("%d%d",&m,&n);
for(int i=1;i<=n;i++)
scanf("%s",mp[i]+1);
memset(dp,127,sizeof(dp));
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
if(mp[i][j]!='C')
continue;
v[i][j]=1;
mp[i][j]='*';
dfs(i,j,0,0);
printf("%d",ans);
return 0;
}
}