D. Gourmet choice
Mr. Apple, a gourmet, works as editor-in-chief of a gastronomic periodical. He travels around the world, tasting new delights of famous chefs from the most fashionable restaurants. Mr. Apple has his own signature method of review — in each restaurant Mr. Apple orders two sets of dishes on two different days. All the dishes are different, because Mr. Apple doesn't like to eat the same food. For each pair of dishes from different days he remembers exactly which was better, or that they were of the same quality. After this the gourmet evaluates each dish with a positive integer.
Once, during a revision of a restaurant of Celtic medieval cuisine named «Poisson», that serves chestnut soup with fir, warm soda bread, spicy lemon pie and other folk food, Mr. Apple was very pleasantly surprised the gourmet with its variety of menu, and hence ordered too much. Now he's confused about evaluating dishes.
The gourmet tasted a set of n
dishes on the first day and a set of m dishes on the second day. He made a table a of size n×m, in which he described his impressions. If, according to the expert, dish i from the first set was better than dish j from the second set, then aij is equal to ">", in the opposite case aij is equal to "<". Dishes also may be equally good, in this case aij
is "=".
Now Mr. Apple wants you to help him to evaluate every dish. Since Mr. Apple is very strict, he will evaluate the dishes so that the maximal number used is as small as possible. But Mr. Apple also is very fair, so he never evaluates the dishes so that it goes against his feelings. In other words, if aij
is "<", then the number assigned to dish i from the first set should be less than the number of dish j from the second set, if aij is ">", then it should be greater, and finally if aij
is "=", then the numbers should be the same.
Help Mr. Apple to evaluate each dish from both sets so that it is consistent with his feelings, or determine that this is impossible.
The first line contains integers n
and m (1≤n,m≤1000
) — the number of dishes in both days.
Each of the next n
lines contains a string of m symbols. The j-th symbol on i-th line is aij
. All strings consist only of "<", ">" and "=".
The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.
If case an answer exist, on the second line print n
integers — evaluations of dishes from the first set, and on the third line print m
integers — evaluations of dishes from the second set.
3 4
>>>>
>>>>
>>>>
Yes
2 2 2
1 1 1 1
3 3
>>>
<<<
>>>
Yes
3 1 3
2 2 2
3 2
==
=<
==
No
In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 2
, for all dishes of the first day.
In the third sample, the table is contradictory — there is no possible evaluation of the dishes that satisfies it.
题意
有两个点集分别有n个点和m个点，告诉你第一个集合里的所有点的权值和第二个集合里的所有点的权值的大小关系。
问是否能给每个点赋一个值，使得满足题目所给的大小关系，如果可以，输出任意一组合法解。
题解：
对于任意两个点ai和bj，如果ai<bj则，连一条从j到i的单向边，问题即为每个点的最长路。
如果相等，我们缩点即可。如果有环则说明无解。
具体步骤：
先缩点，再判环，我直接tarjan求环，然后再在DAG上dfs求最长路即可。
Debug:
１．一开始先不管相等的边，判环再缩点，导致可能缩点之后又有新的环产生。
２．一看n<1000，平时做多了n和边是一个数量级的题目，于是边只开了10000，还以为够了，真是惯性思维惹的祸，还是太菜了。
代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define N 2000050 4 char road[2000][2000]; 5 int n,m,fa[N],f[N],ans[N]; 6 bool flag=0; 7 struct Edge{int from,to,s;}edges[N<<1],edges2[N<<2]; 8 int tot,last[N]; 9 template<typename T>void read(T&x) 10 { 11 int k=0;char c=getchar(); 12 x=0; 13 while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar(); 14 if(c==EOF)exit(0); 15 while(isdigit(c))x=x*10+c-'0',c=getchar(); 16 x=k?-x:x; 17 } 18 void read_char(char &c) 19 {while((c=getchar())!='<'&&c!='>'&&c!='='&&c!=EOF);} 20 int gf(int x) 21 { 22 if (x==fa[x])return x; 23 fa[x]=gf(fa[x]); 24 return fa[x]; 25 } 26 void AddEdge(int x,int y) 27 { 28 edges[++tot]=Edge{x,y,last[x]}; 29 last[x]=tot; 30 } 31 int low[N],dfn[N],sk[N],at_sk[N],top,Time; 32 void tarjan(int x) 33 { 34 dfn[x]=++Time; 35 low[x]=Time; 36 sk[++top]=x; 37 at_sk[x]=1; 38 f[x]=1; 39 for(int i=last[x];i;i=edges[i].s) 40 { 41 Edge &e=edges[i]; 42 if (gf(x)==gf(e.to))flag=true; 43 if (!f[e.to])tarjan(e.to); 44 if (at_sk[e.to])low[x]=min(low[x],low[e.to]); 45 } 46 if (sk[top]!=x)flag=true; 47 if (low[x]==dfn[x]) 48 { 49 while(sk[top+1]!=x&&top) 50 at_sk[sk[top--]]=0; 51 } 52 } 53 void dfs(int x) 54 { 55 f[x]=1; 56 ans[x]=0; 57 for(int i=last[x];i;i=edges[i].s) 58 { 59 Edge &e=edges[i]; 60 if(!ans[e.to]&&!f[e.to])dfs(e.to); 61 ans[x]=max(ans[x],ans[e.to]); 62 } 63 ans[x]++; 64 } 65 int main() 66 { 67 #ifndef ONLINE_JUDGE 68 freopen("aa.in","r",stdin); 69 #endif 70 read(n);read(m); 71 char c; 72 for(int i=1;i<=n+m+n;i++)fa[i]=i; 73 for(int i=1;i<=n;i++) 74 for(int j=1;j<=m;j++) 75 { 76 read_char(c); 77 //if (c=='<')AddEdge(j+n,i); 78 //if (c=='>')AddEdge(i,j+n); 79 road[i][j]=c; 80 if (c=='=')fa[gf(i)]=gf(j+n); 81 } 82 for(int i=1;i<=n;i++) 83 for(int j=1;j<=m;j++) 84 if (road[i][j]=='<')AddEdge(gf(j+n),gf(i)); 85 else if (road[i][j]=='>')AddEdge(gf(i),gf(j+n)); 86 87 for(int i=1;i<=n+m;i++) if(!f[gf(i)]&&!flag)tarjan(gf(i)); 88 if (flag) 89 { 90 printf("No"); 91 return 0; 92 } 93 printf("Yes "); 94 memset(f,0,sizeof(f)); 95 for(int i=1;i<=n+m;i++) if (!f[gf(i)]) dfs(gf(i)); 96 for(int i=1;i<=n;i++) printf("%d ",ans[gf(i)]); 97 putchar(' '); 98 for(int i=1;i<=m;i++) printf("%d ",ans[gf(i+n)]); 99 }