HDU1241Oil Deposits
求聯通塊數量
Sample Input
1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5?
****@
*@@*@
*@**@
@@@*@
@@**@
0 0?
Sample Output
0
1
2
2

DFS:
?

#include <iostream> #include<cstdio> #include <iostream> #include <cstring> #include <queue> using namespace std; typedef long long ll; int n,m; char map[110][110]; int dir[10][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{-1,1},{1,-1},{-1,-1}}; struct node{int x,y;//node(int xx,int yy){x=xx,y=yy;} }; void dfs(int x,int y){for(int i=0;i<8;++i){int dx=x+dir[i][0];int dy=y+dir[i][1];if(dx<1||dx>n||dy<1||dy>m)continue;if(map[dx][dy]=='@'){map[dx][dy]='*';dfs(dx,dy);}} } int main(){while(scanf("%d %d",&n,&m)!=EOF){if(n==0||m==0)break;for(int i=1;i<=n;++i)scanf("%s",map[i]+1);int ans=0;for(int i=1;i<=n;++i)for(int j=1;j<=m;++j){if(map[i][j]=='@'){map[i][j]='*';ans++;dfs(i,j);}}printf("%d\n",ans);}return 0; }

bfs:

#include <iostream>#include<cstdio>#include <iostream>#include <cstring>#include <queue>using namespace std;typedef long long ll;int n,m;char map[110][110];int dir[10][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{-1,1},{1,-1},{-1,-1}};struct node{int x,y;//node(int xx,int yy){x=xx,y=yy;}};void bfs(int x,int y){queue<node> q;node point;point.x=x; point.y=y;q.push(point);while(!q.empty()){//cout<<1<<endl;node cur=q.front();q.pop();map[cur.x][cur.y]='*';node tmp;cout<<cur.x<<' '<<cur.y<<' '<<map[cur.x][cur.y]<<endl;for(int i=0;i<8;++i){int dx=cur.x+dir[i][0];int dy=cur.y+dir[i][1];if(dx<1||dx>n||dy<1||dy>m)continue;if(map[dx][dy]=='@'){map[dx][dy]='*';tmp.x=dx;tmp.y=dy;q.push(tmp);}}}}int main(){while(scanf("%d %d",&n,&m)!=EOF){if(n==0||m==0)break;for(int i=1;i<=n;++i)scanf("%s",map[i]+1);int ans=0;for(int i=1;i<=n;++i)for(int j=1;j<=m;++j){if(map[i][j]=='@'){map[i][j]='*';ans++;bfs(i,j);}}printf("%d\n",ans);}return 0;}

HDU 2 612?Find a way
題意就是在圖中找一條最短路徑使得 Y M到一家kfc(@)約會,求出約會最短的路徑
@可能有多家,#是障礙物
各種Exceeded得姿勢,先是用的pair導致得是TLE，改了之后又MLE,之后又回到TLE，
最終倆個TLE一個是pair還有一個是vis標記得時候在進度隊列的時候進行標記得話，相對來說會減少很多次進隊次數

#include <cstdio> #include <iostream> #include <queue> #include <bits/stdc++.h> #include <cstring> using namespace std; const int inf=99999; int n,m; char mp[201][201]; int step[201][201][2]; int vis[201][201]; int dir[4][2]={{1,0},{0,1},{0,-1},{-1,0}}; int flag; struct node{ int x,y;}; void bfs(int x,int y){queue<node >q;node p,tmp;//pair<int,int> p;p.x=x,p.y=y;q.push(p);while(!q.empty()){p=q.front();q.pop();//int x=p.x;//int y=p.y;//vis[p.x][p.y]=1;for(int i=0;i<4;++i){tmp.x=p.x+dir[i][0];tmp.y=p.y+dir[i][1];if(tmp.x<0||tmp.x>=n||tmp.y<0||tmp.y>=m||vis[tmp.x][tmp.y]||mp[tmp.x][tmp.y]=='#')continue;step[tmp.x][tmp.y][flag]=step[p.x][p.y][flag]+1;q.push(tmp);vis[tmp.x][tmp.y]=1;}} } int main(){while(scanf("%d%d",&n,&m)!=EOF){for(int i=0;i<n;++i)scanf("%s",mp[i]);memset(step, inf,sizeof(step));for(int i=0;i<n;++i)for(int j=0;j<m;++j){if(mp[i][j]=='Y'){memset(vis,0,sizeof(vis));flag=0;step[i][j][flag]=0;bfs(i,j);}if(mp[i][j]=='M'){memset(vis,0,sizeof(vis));flag=1;step[i][j][flag]=0;bfs(i,j);}}int dis=inf;for(int i=0;i<n;++i)for(int j=0;j<m;++j){if(mp[i][j]=='@'&&dis>step[i][j][0]+step[i][j][1]){dis=step[i][j][0]+step[i][j][1];//dis=min(dis,step[i][j][0]+step[i][j][1]);}}printf("%d\n",dis*11);}return 0; } //MLE&&TLE: #include <cstdio> #include <iostream> #include <queue> #include <bits/stdc++.h> #include <cstring> using namespace std; const int inf=0x3f3f3f3f; int n,m; char mp[220][220]; int step[220][220][2]; int vis[220][220]; int dir[4][2]={{1,0},{0,1},{0,-1},{-1,0}}; int flag; void bfs(int x,int y){queue<pair<int,int> >q;pair<int,int> p;q.push(make_pair(x,y));while(!q.empty()){p=q.front();q.pop();int x=p.first;int y=p.second;for(int i=0;i<4;++i){int dx=x+dir[i][0];int dy=y+dir[i][1];if(dx<1||dx>n||dy<1||dy>m||vis[dx][dy]||mp[dx][dy]=='#')continue;step[dx][dy][flag]=step[x][y][flag]+1;q.push(make_pair(dx,dy));vis[x][y]=1;}} } int main(){while(scanf("%d%d",&n,&m)!=EOF){for(int i=1;i<=n;++i)scanf("%s",mp[i]+1);memset(step, inf,sizeof(step));for(int i=1;i<=n;++i)for(int j=1;j<=m;++j)step[i][j][0]=step[i][j][1]=inf;for(int i=1;i<=n;++i)for(int j=1;j<=m;++j){if(mp[i][j]=='Y'){memset(vis,0,sizeof(vis));flag=0;step[i][j][flag]=0;bfs(i,j);}if(mp[i][j]=='M'){memset(vis,0,sizeof(vis));flag=1;step[i][j][flag]=0;bfs(i,j);}}int dis=inf;for(int i=1;i<=n;++i)for(int j=1;j<=m;++j){if(mp[i][j]=='@'){dis=min(dis,step[i][j][0]+step[i][j][1]);}}printf("%d\n",dis*11);}return 0; }

?

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