poj 1128
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这道题是一道应用拓扑排序算法的题目,不过要注意如果有多种情况,需要按字典序全部输出。输出所有拓扑排序的方案的方法是深搜入度为0的节点。
代码(C++):
#include <iostream>#include <cstdio>#include <set>#include <algorithm>#include <cstring>#include <string>#define MAX_N 35#define MAX_M 100#define INF (1<<30)using namespace std;struct Edge{ int v; int next;} edge[MAX_M];set<char> sc;int h,w,num,head[MAX_N],indeg[MAX_N];char frame[MAX_N][MAX_N];bool vis[MAX_N],g[MAX_N][MAX_N],exist[MAX_N];void add_edge(int u,int v){ if(g[u][v]==true) return; g[u][v]=true; indeg[v]++; edge[num].v=v; edge[num].next=head[u]; head[u]=num; num++;}void build_g(){ set<char>::iterator si; int i,j,a,b,c,d; char tmp; num=0; memset(head,-1,sizeof(head)); memset(indeg,0,sizeof(indeg)); memset(g,false,sizeof(g)); memset(exist,false,sizeof(exist)); for(si=sc.begin();si!=sc.end();si++) { tmp=*si; exist[tmp-'A']=true; a=c=INF; b=d=-1; for(i=0;i<h;i++) { for(j=0;j<w;j++) { if(frame[i][j]==tmp) { a=min(a,j); b=max(b,j); c=min(c,i); d=max(d,i); } } } for(i=c;i<=d;i++) { if(frame[i][a]!=tmp) add_edge(tmp-'A',frame[i][a]-'A'); if(frame[i][b]!=tmp) add_edge(tmp-'A',frame[i][b]-'A'); } for(i=a+1;i<b;i++) { if(frame[c][i]!=tmp) add_edge(tmp-'A',frame[c][i]-'A'); if(frame[d][i]!=tmp) add_edge(tmp-'A',frame[d][i]-'A'); } }}void dfs(int u,string s){ int i; s+=char(u+'A'); if(s.length()==sc.size()) { printf("%s\n",s.c_str()); return; } vis[u]=true; for(i=head[u];i!=-1;i=edge[i].next) { indeg[edge[i].v]--; } for(i=0;i<26;i++) { if(exist[i]&&indeg[i]==0&&!vis[i]) { dfs(i,s); } } vis[u]=false; for(i=head[u];i!=-1;i=edge[i].next) { indeg[edge[i].v]++; }}void toposort(){ int i; memset(vis,false,sizeof(vis)); for(i=0;i<26;i++) { if(exist[i]&&indeg[i]==0) dfs(i,""); }}int main(){ //freopen("in.txt","r",stdin); int i,j; while(scanf("%d",&h)!=EOF) { scanf("%d",&w); for(i=0;i<h;i++) scanf("%s",frame[i]); for(i=0;i<h;i++) { for(j=0;j<w;j++) { if(frame[i][j]!='.') sc.insert(frame[i][j]); } } build_g(); toposort(); sc.clear(); } return 0;}
题目(http://poj.org/problem?id=1128):
Frame Stacking
Time Limit: 1000MS Memory Limit: 10000K
Description
Consider the following 5 picture frames placed on an 9 x 8 array.
Now place them on top of one another starting with 1 at the bottom and ending up with 5 on top. If any part of a frame covers another it hides that part of the frame below.
Viewing the stack of 5 frames we see the following.
In what order are the frames stacked from bottom to top? The answer is EDABC.
Your problem is to determine the order in which the frames are stacked from bottom to top given a picture of the stacked frames. Here are the rules:
1. The width of the frame is always exactly 1 character and the sides are never shorter than 3 characters.
2. It is possible to see at least one part of each of the four sides of a frame. A corner shows two sides.
3. The frames will be lettered with capital letters, and no two frames will be assigned the same letter.
........ ........ ........ ........ .CCC....EEEEEE.. ........ ........ ..BBBB.. .C.C....E....E.. DDDDDD.. ........ ..B..B.. .C.C....E....E.. D....D.. ........ ..B..B.. .CCC....E....E.. D....D.. ....AAAA ..B..B.. ........E....E.. D....D.. ....A..A ..BBBB.. ........E....E.. DDDDDD.. ....A..A ........ ........E....E.. ........ ....AAAA ........ ........EEEEEE.. ........ ........ ........ ........ 1 2 3 4 5
Now place them on top of one another starting with 1 at the bottom and ending up with 5 on top. If any part of a frame covers another it hides that part of the frame below.
Viewing the stack of 5 frames we see the following.
.CCC....ECBCBB..DCBCDB..DCCC.B..D.B.ABAAD.BBBB.ADDDDAD.AE...AAAAEEEEEE..
In what order are the frames stacked from bottom to top? The answer is EDABC.
Your problem is to determine the order in which the frames are stacked from bottom to top given a picture of the stacked frames. Here are the rules:
1. The width of the frame is always exactly 1 character and the sides are never shorter than 3 characters.
2. It is possible to see at least one part of each of the four sides of a frame. A corner shows two sides.
3. The frames will be lettered with capital letters, and no two frames will be assigned the same letter.
Input
Each input block contains the height, h (h<=30) on the first line and the width w (w<=30) on the second. A picture of the stacked frames is then given as h strings with w characters each.
Your input may contain multiple blocks of the format described above, without any blank lines in between. All blocks in the input must be processed sequentially.
Your input may contain multiple blocks of the format described above, without any blank lines in between. All blocks in the input must be processed sequentially.
Output
Write the solution to the standard output. Give the letters of the frames in the order they were stacked from bottom to top. If there are multiple possibilities for an ordering, list all such possibilities in alphabetical order, each one on a separate line. There will always be at least one legal ordering for each input block. List the output for all blocks in the input sequentially, without any blank lines (not even between blocks).
Sample Input
98.CCC....ECBCBB..DCBCDB..DCCC.B..D.B.ABAAD.BBBB.ADDDDAD.AE...AAAAEEEEEE..
Sample Output
EDABC
0 0
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