标签:des blog http io ar os sp for java
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 621 Accepted Submission(s): 254
Case n: Island splits when ocean rises f feet.or
Case n: Island never splits.Our convention here is if your answer is, say, 5 feet, you more accurately mean “5 feet plus a little more.” That is, at least a little water will be flowing over the originally 5 foot high portion of land.
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<string>
#include<queue>
#include<vector>
#include<set>
using namespace std;
const int MAX=105;
int grid[MAX][MAX];
bool vis[MAX][MAX];
int d[4][2]={{-1,0},{1,0},{0,1},{0,-1}};
int n,m;
int max(int a,int b)
{
return a>b?a:b;
}
void dfs(int i,int j,int h)
{
if(i<1||j<1||i>n||j>m||vis[i][j]||grid[i][j]>h)
return;
vis[i][j]=true;
grid[i][j]=0;
for(int k=0;k<4;k++)
dfs(i+d[k][0],j+d[k][1],h);
}
void dfs2(int i,int j)
{
if(i<1||j<1||i>n||j>m||vis[i][j]||(!grid[i][j]))
return;
vis[i][j]=true;
for(int k=0;k<4;k++)
dfs2(i+d[k][0],j+d[k][1]);
}
bool check(int h)
{
int i,j,cnt=0;
memset(vis,0,sizeof(vis));
for(i=1;i<=m;i++)
if(!vis[1][i])
dfs(1,i,h);
for(i=1;i<=n;i++)
if(!vis[i][1])
dfs(i,1,h);
for(i=1;i<=m;i++)
if(!vis[n][i])
dfs(n,i,h);
for(i=1;i<=n;i++)
if(!vis[i][m])
dfs(i,m,h);
memset(vis,0,sizeof(vis));
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
if(!vis[i][j]&&grid[i][j])
{
cnt++;
dfs2(i,j);
}
if(cnt<=1)
return 1;
else
return 0;
}
int main()
{
int ca=0,i,j,maxx=-1,cnt,ans;
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==0&&m==0)
break;
cnt=0;
ans=-1;
memset(grid,0,sizeof(grid));
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
{
scanf("%d",&grid[i][j]);
maxx=max(maxx,grid[i][j]);
}
for(i=1;i<=maxx;i++)
if(!check(i))
{
ans=i;
break;
}
printf("Case %d: ",++ca);
if(ans==-1)
printf("Island never splits.\n");
else
printf("Island splits when ocean rises %d feet.\n",ans);
}
return 0;
}
标签:des blog http io ar os sp for java
原文地址:http://www.cnblogs.com/a972290869/p/4164032.html