D. Gourmet choice并查集，拓扑结构

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D. Gourmet choice
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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 nn dishes on the first day and a set of mm dishes on the second day. He made a table aa of size n×mn×m, in which he described his impressions. If, according to the expert, dish ii from the first set was better than dish jj from the second set, then aijaij is equal to ">", in the opposite case aijaij is equal to "<". Dishes also may be equally good, in this case aijaij 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 aijaij is "<", then the number assigned to dish ii from the first set should be less than the number of dish jj from the second set, if aijaij is ">", then it should be greater, and finally if aijaij 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.

Input

The first line contains integers nn and mm (1≤n,m≤10001≤n,m≤1000) — the number of dishes in both days.

Each of the next nn lines contains a string of mm symbols. The jj-th symbol on ii-th line is aijaij. All strings consist only of "<", ">" and "=".

Output

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 nn integers — evaluations of dishes from the first set, and on the third line print mmintegers — evaluations of dishes from the second set.

Examples
input
Copy
3 4
>>>>
>>>>
>>>>
output
Copy
Yes
2 2 2 
1 1 1 1
input
Copy
3 3
>>>
<<<
>>>
output
Copy
Yes
3 1 3 
2 2 2
input
Copy
3 2
==
=<
==
output
Copy
No
Note

In the first sample, all dishes of the first day are better than dishes of the second day. So, the highest score will be 22, 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：问有没有符合条件的n个数和m个数满足矩阵的条件

以下是大神的解答，我还不太会

题解：一看有比较大小，很明显我们可以通过拓扑排序来完成，小的指向大的，每一个数取符合条件的最小的（即指向他的最大的数+1）即可，至于相等的呢，我们就把他们先连接到一块，傻逼了，写错了个地方，并查集连接到一块的只有祖先有值，写成了取父亲的，也是操蛋，。。。。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=150010;
int a[2100];
int pre[2100];
int n,m;
char s[1100][1100];
vector<int> v[2100];
int in[2100];
int f(int x)
{
    return x==pre[x]?x:pre[x]=f(pre[x]);
}
int main()
{
    int ans=0;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=2000;i++) pre[i]=i; 
    for(int i=1;i<=n;i++)
    {
        scanf("%s",s[i]+1);
        for(int j=1;j<=m;j++)
        {
            if(s[i][j]==‘=‘)
            {
                int xx=f(i);
                int yy=f(j+1000);
                if(xx==yy) continue;
                pre[yy]=xx;
                ans++;
            }
        }
    }
 
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(s[i][j]==‘=‘) continue;
            int xx=f(i);
            int yy=f(j+1000);
            
            if(s[i][j]==‘<‘)
            {
                v[xx].push_back(yy);
                in[yy]++;
            }
            else
            {
                v[yy].push_back(xx);
                in[xx]++;
            }
        }
    }
    queue<int> q;
    for(int i=1;i<=n;i++)
    {
        if(pre[i]==i&&in[i]==0)
        {
            q.push(i);
            a[i]=1;
            ans++;
        }
    }
    for(int i=1001;i<=m+1000;i++)
    {
        if(pre[i]==i&&in[i]==0)
        {
            q.push(i);
            a[i]=1;
            ans++;
        }
    }
    if(q.empty()) printf("No
");
    else
    {
        while(!q.empty())
        {
            int u=q.front();q.pop();
            for(int i=0;i<v[u].size();i++)
            {
                int to=v[u][i];
                in[to]--;
                if(a[to]==0) a[to]=a[u]+1;
                else a[to]=max(a[to],a[u]+1);
                if(in[to]==0) 
                {
                    ans++;q.push(to);
                }
            }
        }
        if(ans!=n+m)
        {
            printf("No
");
            return 0;
        }
        printf("Yes
");
        for(int i=1;i<=n;i++)
        {
        //    cout<<pre[i]<<endl;
            if(pre[i]!=i) a[i]=a[f(i)];
            
            printf("%d%c",a[i]," 
"[i==n]);
        }
        for(int i=1001;i<=m+1000;i++)
        {
            if(pre[i]!=i) a[i]=a[f(i)];
            
            printf("%d%c",a[i]," 
"[i==m+1000]);
        }
    }
    return 0;
}

以下是cf其他的解答

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue> 

using namespace std;
const int N=2005;

vector<int> v[N];
queue<int> Q; 
int n,m,col[N],stk[N],top,cnt,val[N][N],zz[N];
int dfn[N],low[N],ans[N],col_num,f[N],num,d[N];
bool vis[N];
struct Edge{
    int u,v,w;    
}edge[N*N];

inline void add(int x,int y,int z){
    v[x].push_back(y); val[x][y]=z;
}

void tarjan(int x){
    dfn[x]=low[x]=++num;
    stk[++top]=x; vis[x]=1; int u;
    for(int i=0;i<v[x].size();i++){
        u=v[x][i];
        if(!dfn[u]) tarjan(u),low[x]=min(low[x],low[u]);
        else if(vis[u]) low[x]=min(low[x],dfn[u]);
    }
    if(dfn[x]!=low[x]) return;
    col[x]=++col_num; vis[x]=0; int lst=x;
    while(stk[top]!=x){
        col[stk[top]]=col_num;
        vis[stk[top]]=0;
        zz[col_num]+=val[stk[top]][lst];
        lst=stk[top]; top--;
    } zz[col_num]+=val[lst][x]; top--;
}

int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++){
            char ch=getchar();
            while(ch!=‘=‘ && ch!=‘<‘ && ch!=‘>‘) ch=getchar();
            if(ch==‘<‘){
                add(i,j+n,1);
                edge[++cnt].u=i,edge[cnt].v=j+n,edge[cnt].w=1;
            }
            else if(ch==‘>‘){
                add(j+n,i,1);
                edge[++cnt].u=j+n,edge[cnt].v=i,edge[cnt].w=1;    
            }
            else {
                edge[++cnt].u=i; edge[cnt].v=j+n; edge[cnt].w=0;
                edge[++cnt].u=j+n; edge[cnt].v=i; edge[cnt].w=0;
                add(i,j+n,0); add(j+n,i,0);    
            }
        }
    for(int i=1;i<=n+m;i++) if(!dfn[i]) tarjan(i);
    for(int i=1;i<=col_num;i++)
        if(zz[i]>0) {puts("No"); return 0;} 
    for(int i=1;i<=col_num;i++)
        v[i].clear();
    memset(val,0,sizeof(val));
    for(int i=1;i<=cnt;i++){
        int u=edge[i].u,v=edge[i].v,w=edge[i].w;
        if(col[u]==col[v]) continue;
        add(col[u],col[v],w); d[col[v]]++;
    }
    for(int i=1;i<=col_num;i++)
        if(!d[i]) f[i]=1,Q.push(i);
    while(Q.size()){
        int x=Q.front(); Q.pop();
        for(int i=0;i<v[x].size();i++){
            int u=v[x][i];
            f[u]=max(f[u],f[x]+val[x][u]);
            d[u]--; if(!d[u]) Q.push(u);    
        }
    }
    puts("Yes");
    for(int i=1;i<=n;i++) printf("%d ",f[col[i]]);
    puts("");
    for(int i=n+1;i<=n+m;i++) printf("%d ",f[col[i]]);
    return 0;    
}

 

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