codeforces #541 D. Gourmet choice(拓扑+并查集)
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.
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 "=".
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.
3 4
>>>>
>>>>
>>>>
Yes
2 2 2
1 1 1 1
3 3
>>>
<<<
>>>
Yes
3 1 3
2 2 2
3 2
==
=<
==
第一天有n个菜,第二天有m个菜。
给一个n*m矩阵,含有(或=、<),a(i,j)指的是:第一天的第i件菜的美味度>(或=、<)第二天的第j件菜的美味度。求出满足这个矩阵的n+m件菜的各自的美味度,要求使用的最大数字尽量小。
思路:如果是等号就用并查集连在一起(缩点) 其他的就拓扑排序
#include <cstdio>
#include <map>
#include <iostream>
#include<cstring>
#include<bits/stdc++.h>
#define ll long long int
#define M 6
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[]={,,,,,,,,,,,,};
int dir[][]={, ,, ,-, ,,-};
int dirs[][]={, ,, ,-, ,,-, -,- ,-, ,,- ,,};
const int inf=0x3f3f3f3f;
const ll mod=1e9+;
int n,m;
int f[];
char G[][];
vector<int> v[];
int ru[];
int ans[];
bool flag;
int find(int x){
if(x!=f[x])
f[x]=find(f[x]);
return f[x];
}
void join(int x,int y){
int xx=find(x); int yy=find(y);
if(xx!=yy){
f[yy]=xx;
}
}
void toposort(){
queue<pair<int,int> > q;
for(int i=;i<=n+m;i++)
if(!ru[i]&&i==find(i)) q.push(make_pair(i,));
while(!q.empty()){
int x=q.front().first;
int y=q.front().second;
q.pop();
ans[x]=y;
int len=v[x].size();
for(int i=;i<len;i++){
int to=find(v[x][i]);
ru[to]--;
if(!ru[to]){
q.push(make_pair(to,y+));
}
}
}
}
int main(){
ios::sync_with_stdio(false);
cin>>n>>m;
for(int i=;i<=n+m;i++) f[i]=i;
flag=;
for(int i=;i<=n;i++)
for(int j=;j<=m;j++){
cin>>G[i][j];
if(G[i][j]=='=') //合并
join(i,j+n);
}
for(int i=;i<=n;i++)
for(int j=;j<=m;j++){
if(G[i][j]=='=') continue;
int xx=find(i); int yy=find(n+j);
if(xx==yy){
flag=;
break;
}
if(G[i][j]=='>') v[yy].push_back(xx),ru[xx]++; //计算入度
if(G[i][j]=='<') v[xx].push_back(yy),ru[yy]++;
if(!flag) break;
} if(!flag) cout<<"No"<<endl;
else{
toposort();
for(int i=;i<=n+m;i++)
if(ans[find(i)]==){
flag=;
break;
}
if(!flag) cout<<"No"<<endl;
else{
cout<<"Yes"<<endl;
for(int i=;i<=n;i++)
cout<<ans[find(i)]<<" ";
cout<<endl;
for(int i=;i<=m;i++)
cout<<ans[find(n+i)]<<" ";
cout<<endl;
}
}
return ;
}
codeforces #541 D. Gourmet choice(拓扑+并查集)的更多相关文章
- CF1131D Gourmet choice(并查集,拓扑排序)
这题CF给的难度是2000,但我感觉没这么高啊…… 题目链接:CF原网 题目大意:有两个正整数序列 $a,b$,长度分别为 $n,m$.给出所有 $a_i$ 和 $b_j(1\le i\le n,1\ ...
- codeforces #541 F Asya And Kittens(并查集+输出路径)
F. Asya And Kittens Asya loves animals very much. Recently, she purchased nn kittens, enumerated the ...
- Codeforces Round #541 (Div. 2) D(并查集+拓扑排序) F (并查集)
D. Gourmet choice 链接:http://codeforces.com/contest/1131/problem/D 思路: = 的情况我们用并查集把他们扔到一个集合,然后根据 > ...
- Codeforces Round #376 (Div. 2) C. Socks---并查集+贪心
题目链接:http://codeforces.com/problemset/problem/731/C 题意:有n只袜子,每只都有一个颜色,现在他的妈妈要去出差m天,然后让他每天穿第 L 和第 R 只 ...
- Codeforces 766D. Mahmoud and a Dictionary 并查集 二元敌对关系 点拆分
D. Mahmoud and a Dictionary time limit per test:4 seconds memory limit per test:256 megabytes input: ...
- CodeForces Roads not only in Berland(并查集)
H - Roads not only in Berland Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d ...
- Codeforces #345div1 C Table Compression (650C) 并查集
题意:给你一个n*m的矩阵,需要在不改变每一行和每一列的大小关系的情况下压缩一个矩阵,压缩后的矩阵所有数的总和尽量的小. 思路:我们有这样的初步设想:对于在一行或一列的数x,y,若x<y,则建立 ...
- hdu 1811 Rank of Tetris (拓扑 & 并查集)
Rank of Tetris Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)To ...
- NOIp 2015信息传递【tarjan/拓扑/并查集】
一道好的NOIp题目,在赛场上总能用许多算法A掉.比如这道和关押罪犯. 题目传送门 法一:tarjan在有向图中跑最小环 有人从别人口中得知自己信息,等效于出现了一个环.于是 这就变成了一个有向图ta ...
随机推荐
- laravel打印sql
DB::connection()->enableQueryLog(); print_r(DB::getQueryLog());
- 6-2 Verbs and Adjectives with that clauses
1 Many sentences in English contain two clauses: a main clause and a "that" clause. The &q ...
- day 7-20 视图,触发器,事务
一.视图 视图是一个虚拟表(非真实存在),其本质是[根据SQL语句获取动态的数据集,并为其命名],用户使用时只需使用[名称]即可获取结果集,可以将该结果集当做表来使用. 使用视图我们可以把查询过程中的 ...
- python3 阿里云控制SLB权重
一.配置好RAM账号的权限(SLB管理权限) 二.安装依赖 pip3 install aliyun-python-sdk-slb pip3 install aliyun-python-sdk-core ...
- IBM rational rose画时序图软件破解安装
上边这个链接是开头的安装步骤,照着链接中的步骤安装完之后,接下来看下边. 1.然后安装完成打开软件“IBM Rational License Keyadministrator”.出现下图:选中第二项“ ...
- python 三目运算符
格式: true_res if condition else false_res Meto 1: Meto 2: >>> x = 2 >>> x+1 if x!=1 ...
- 三、ASP.NET Core 部署Linux
预备工作 1.删除dotnet core sdk sudo yum erase libunwind libicu 2.删除链接 sudo rm -rf /usr/local/bin 3.sudo yu ...
- Calendar用法随笔
平时在处理时间问题的时候,一般会想到用java.util.Date类型,在使用倒时间的运算的时候,就不是很方便,找找到了java.util.Calendar类,中文意思是“日历”,以下就是自己对这个类 ...
- Memcached cas 陷阱
本地使用的 php7环境,测试好上传到服务器后发现memcached get 报错,服务器上是php5环境: 出错代码如下: $memConnect->get($key,null, Memcac ...
- 在web-inf外面 使用的是绝对路径进行访问 “/”表示访问文件夹 一层一层方式 我们在windos下访问文件夹也是一层一层的访问