Cow Ski Area
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 3323 | Accepted: 919 |
Description
Farmer John‘s cousin, Farmer Ron, who lives in the mountains of Colorado, has recently taught his cows to ski. Unfortunately, his cows are somewhat timid and are afraid to ski among crowds of people at the local resorts, so FR has decided to construct his own private ski area behind his farm.
FR‘s ski area is a rectangle of width W and length L of ‘land
squares‘ (1 <= W <= 500; 1 <= L <= 500). Each land square
is an integral height H above sea level (0 <= H <= 9,999). Cows
can ski horizontally and vertically between any two adjacent land
squares, but never diagonally. Cows can ski from a higher square to a
lower square but not the other way and they can ski either direction
between two adjacent squares of the same height.
FR wants to build his ski area so that his cows can travel between
any two squares by a combination of skiing (as described above) and ski
lifts. A ski lift can be built between any two squares of the ski area,
regardless of height. Ski lifts are bidirectional. Ski lifts can cross
over each other since they can be built at varying heights above the
ground, and multiple ski lifts can begin or end at the same square.
Since ski lifts are expensive to build, FR wants to minimize the number
of ski lifts he has to build to allow his cows to travel between all
squares of his ski area.
Find the minimum number of ski lifts required to ensure the cows can
travel from any square to any other square via a combination of skiing
and lifts.
Input
* Line 1: Two space-separated integers: W and L
* Lines 2..L+1: L lines, each with W space-separated integers corresponding to the height of each square of land.
Output
*
Line 1: A single integer equal to the minimal number of ski lifts FR
needs to build to ensure that his cows can travel from any square to any
other square via a combination of skiing and ski lifts
Sample Input
9 3 1 1 1 2 2 2 1 1 1 1 2 1 2 3 2 1 2 1 1 1 1 2 2 2 1 1 1
Sample Output
3SB题，还花了好长时间，不开心，不写题解了。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define inf 0x3f3f3f3f
#define mod 10000
typedef long long ll;
using namespace std;
const int N=505;
const int M=250005;
int s,t,n,m,cnt,tim,top,cut,k;
int head[M],dfn[M],low[M],stack1[M];
int num[M],in[M],out[M],vis[M],w[N][N];
int dis[4][2]= {0,1,1,0,-1,0,0,-1};
bool flag=false;
struct man {
int to,nxt;
} edg[M*10];
void addedg(int u,int v) {
edg[cnt].to=v;
edg[cnt].nxt=head[u];
head[u]=cnt++;//printf("!!!%d %d\n",u,v);system("pause");
}
void init() {
cnt=0;
tim=0;
top=cut=k=0;
memset(head,-1,sizeof head);
memset(dfn,0,sizeof dfn);
memset(low,0,sizeof low);
memset(stack1,0,sizeof stack1);
memset(num,0,sizeof num);
memset(in,0,sizeof in);
memset(out,0,sizeof out);
memset(vis,0,sizeof vis);
memset(edg,0,sizeof edg);
memset(w,0,sizeof w);
}
void Tarjan(int u) {
int v;
low[u] = dfn[u] = ++tim;
stack1[top++] = u;
vis[u] = 1;
for(int e = head[u]; e != -1; e = edg[e].nxt)
{
v = edg[e].to;
if(!dfn[v])
{
Tarjan(v);
low[u] = min(low[u], low[v]);
}
else if(vis[v])
{
low[u] = min(low[u], dfn[v]);
}
}
if(low[u] == dfn[u])
{
cut++;
do
{
v = stack1[--top];
num[v] = cut;
vis[v] = 0;
}while(u != v);
}
}
void build(int i,int j,int d)
{
int xx=i+dis[d][0];
int yy=j+dis[d][1];
int u=i*m+j,v=xx*m+yy;
if(xx>=0&&yy<m&&yy>=0&&xx<n){
if(w[i][j]>=w[xx][yy])addedg(u,v);
if(w[i][j]<=w[xx][yy])addedg(v,u);
}
return;
}
int main() {
while(~scanf("%d%d",&m,&n)) {
init();
for(int i=0; i<n; i++) {
for(int j=0; j<m; j++) {
scanf("%d",&w[i][j]);
}
}
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
for(int d = 0; d < 2; d++) {
build(i,j,d);
}
}
}
for(int i=0; i<n*m; i++)if(!dfn[i])Tarjan(i);
for(int i=0; i<n*m; i++) {
for(int j=head[i]; j!=-1; j=edg[j].nxt) {
int v=edg[j].to;
if(num[i]!=num[v])out[num[i]]++,in[num[v]]++;
}
}
int father=0,son=0;
for(int i=1; i<=cut; i++) {
if(in[i]==0)father++;
if(out[i]==0)son++;
}
if(cut==1)printf("0\n");
else printf("%d\n",max(father,son));
}
return 0;
}