URAL 1035 Cross-stitch

A graph of the simulation questions,,,,,//So you're a jerk,,,,, Euler path AH!!!!!!!!

The slash represents the seam line, each lattice vertex can want to do is to have a small hole can wear the past,, and then the idea of the normal stitches,,, asked to sew out the pattern of the minimum number of steps,,,,, sew over, sew the past ~

The beginning is to fully simulate (that is, a situation to write a function), but once through a face, there are 4 situations ...

The result is 200+ line, but still wrong ...

Give a few sets of examples

5 4
.. \/
\/..
/x..
.. x/
.. X.
.. X.
\/\/
\/..
\/..
\/..
Ans:13
3 3
.. \
/\x
...
.. \
///
...
Ans:2

I think a good solution is to ask for the degrees and degrees of each point and then see how many odd ... Have not been verified.

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <bits/stdc++.h>
using namespace Std;

int p[40500],a[40500][2],sum[40500],n,m;

int find (int x)
{
if (p[x]!=x) P[x]=find (p[x]);
return p[x];
}
int judge (int x,int y)
{
Return (Find (x) ==find (y))? 1:0;
}
void make (int side)
{
int ul,ur,dl,dr;
char c;
for (int i=0;i<n;i++)
for (int j=0;j<m;j++)
{
cin>>c;
ul=i* (m+1) +j;ur=ul+1;
Dl= (i+1) * (m+1) +j;dr=dl+1;
if ((c== ' X ') | | (c== ' \ \ '))
{
a[ul][side]++;a[dr][side]++;
if (!judge (UL,DR)) P[P[UL]]=P[DR];
}
if ((c== ' X ') | | (c== '/'))
{
a[ur][side]++;a[dl][side]++;
if (!judge (UR,DL)) P[P[UR]]=P[DL];
}
}
}

int main ()
{
scanf ("%d%d", &n,&m);
int i,j,ans=0;
For (i=0;i< (m+1) * (n+1); i++) p[i]=i;
Make (0);
Make (1);
For (i=0;i< (m+1) * (n+1); i++)
{
if (a[i][0]| | A[I][1]) sum[find (i)]+=abs (a[i][0]-a[i][1]);
}
For (i=0;i< (m+1) * (n+1); i++)
{
if (a[i][0]| | A[I][1]) && (p[i]==i)) ans+=sum[i]?sum[i]/2:1;
}
printf ("%d\n", ans);
return 0;
}

In fact, it's better to use DFS,,, to merge the situation together,,,,,,

Hey



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URAL 1035 Cross-stitch