Codevs 1066 Diversion into the city

My mom, this card has me for two days.

In fact very simple, on each starting point down the BFS, control the continuous section of the arid zone. It is then converted to a DP problem that covers all areas with the fewest segments.

However?!! I didn't know how to write a magical greed at first. The miserable world.

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace Std;
const int maxn=505;
struct SEG
{
int left,right;
}S[MAXN*3];
int map[maxn][maxn],n,m;
int DX[]={0,0,1,0,-1},DY[]={0,-1,0,1,0},CNT=0,DP[MAXN];
BOOL VIS[MAXN][MAXN],K[MAXN];
Queue <int> q;
BOOL CMP (SEG x,seg y)
{
if (x.left==y.left)
Return x.right>y.right;
Return x.left<y.left;
}
BOOL Judge (int x,int y)
{
if ((x>=1) && (x<=n) && (y>=1) && (y<=m) && (Vis[x][y]==false))
return true;
return false;
}
void BFs (int x)
{
memset (vis,false,sizeof (VIS));
while (!q.empty ())
Q.pop ();
int minn=99999999,maxn=0;
if (n==1)
{
Minn=min (MINN,X);
Maxn=max (MAXN,X);
K[x]=true;
}
Vis[1][x]=true;
Q.push (x);
while (!q.empty ())
{
int Head=q.front ();
Q.pop ();
int nx,ny,tx,ty;
if (head%m==0) nx=head/m;
else nx=head/m+1;
if (head-nx*m==0) ny=m;
else ny=head-(nx-1) *m;
for (int i=1;i<=4;i++)
{
int tx=nx+dx[i],ty=ny+dy[i];
if ((Judge (Tx,ty) ==true) && (Map[nx][ny]>map[tx][ty]))
{
Q.push ((tx-1) *m+ty);
Vis[tx][ty]=true;
if (tx==n)
{
Minn=min (Minn,ty);
Maxn=max (Maxn,ty);
K[ty]=true;
}
}
}
}
cnt++;
S[cnt].left=minn;
S[CNT].RIGHT=MAXN;
}
int main ()
{
Memset (K,false,sizeof (k));
Memset (Dp,0x3f,sizeof (DP));
scanf ("%d%d", &n,&m);
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
scanf ("%d", &map[i][j]);
for (int i=1;i<=m;i++)
if ((Map[1][i]>=map[1][i-1]) && (map[1][i]>=map[1][i+1]))
BFS (i);
int num=0;
for (int i=1;i<=m;i++)
if (K[i]==false)
num++;
if (num>0)
{
printf ("0\n%d\n", num);
return 0;
}
Sort (s+1,s+cnt+1,cmp);
dp[0]=0;
for (int i=1;i<=m;i++)
for (int j=1;j<=cnt;j++)
if ((s[j].left<=i) && (s[j].right>=i))
Dp[i]=min (dp[i],dp[s[j].left-1]+1);
printf ("1\n%d\n", Dp[m]);
return 0;
}

Codevs 1066 Diversion into the city