A14263 | Graph Coloring
时间限制1s
内存限制256MB
通过 / 提交0/0
题目描述
You are given a bipartite graph consisting of $n_1$ vertices in the first part, $n_2$ vertices in the second part, and $m$ edges, numbered from $1$ to $m$ . You have to color each edge into one of two colors, red and blue. You have to minimize the following value: $\sum \limits_{v \in V} |r(v) - b(v)|$ , where $V$ is the set of vertices of the graph, $r(v)$ is the number of red edges incident to $v$ , and $b(v)$ is the number of blue edges incident to $v$ .
Sounds classical and easy, right? Well, you have to process $q$ queries of the following format:
- $1$ $v_1$ $v_2$ — add a new edge connecting the vertex $v_1$ of the first part with the vertex $v_2$ of the second part. This edge gets a new index as follows: the first added edge gets the index $m + 1$ , the second — $m + 2$ , and so on. After adding the edge, you have to print the hash of the current optimal coloring (if there are multiple optimal colorings, print the hash of any of them). Actually, this hash won't be verified, you may print any number as the answer to this query, but you may be asked to produce the coloring having this hash;
- $2$ — print the optimal coloring of the graph with the same hash you printed while processing the previous query. The query of this type will only be asked after a query of type $1$ , and there will be at most $10$ queries of this type. If there are multiple optimal colorings corresponding to this hash, print any of them.
Note that if an edge was red or blue in some coloring, it may change its color in next colorings.
The hash of the coloring is calculated as follows: let $R$ be the set of indices of red edges, then the hash is $(\sum \limits_{i \in R} 2^i) \bmod 998244353$ .
Note that you should solve the problem in online mode. It means that you can't read the whole input at once. You can read each query only after writing the answer for the last query. Use functions fflush in C++ and BufferedWriter.flush in Java languages after each writing in your program.
Sounds classical and easy, right? Well, you have to process $q$ queries of the following format:
- $1$ $v_1$ $v_2$ — add a new edge connecting the vertex $v_1$ of the first part with the vertex $v_2$ of the second part. This edge gets a new index as follows: the first added edge gets the index $m + 1$ , the second — $m + 2$ , and so on. After adding the edge, you have to print the hash of the current optimal coloring (if there are multiple optimal colorings, print the hash of any of them). Actually, this hash won't be verified, you may print any number as the answer to this query, but you may be asked to produce the coloring having this hash;
- $2$ — print the optimal coloring of the graph with the same hash you printed while processing the previous query. The query of this type will only be asked after a query of type $1$ , and there will be at most $10$ queries of this type. If there are multiple optimal colorings corresponding to this hash, print any of them.
Note that if an edge was red or blue in some coloring, it may change its color in next colorings.
The hash of the coloring is calculated as follows: let $R$ be the set of indices of red edges, then the hash is $(\sum \limits_{i \in R} 2^i) \bmod 998244353$ .
Note that you should solve the problem in online mode. It means that you can't read the whole input at once. You can read each query only after writing the answer for the last query. Use functions fflush in C++ and BufferedWriter.flush in Java languages after each writing in your program.
输入格式
The first line contains three integers $n_1$ , $n_2$ and $m$ ( $1 \le n_1, n_2, m \le 2 \cdot 10^5$ ).
Then $m$ lines follow, the $i$ -th of them contains two integers $x_i$ and $y_i$ ( $1 \le x_i \le n_1$ ; $1 \le y_i \le n_2$ ) meaning that the $i$ -th edge connects the vertex $x_i$ from the first part and the vertex $y_i$ from the second part.
The next line contains one integer $q$ ( $1 \le q \le 2 \cdot 10^5$ ) — the number of queries you have to process.
The next $q$ lines contain the queries in the format introduced in the statement.
Additional constraints on the input:
- at any moment, the graph won't contain any multiple edges;
- the queries of type $2$ are only asked if the previous query had type $1$ ;
- there are at most $10$ queries of type $2$ .
Then $m$ lines follow, the $i$ -th of them contains two integers $x_i$ and $y_i$ ( $1 \le x_i \le n_1$ ; $1 \le y_i \le n_2$ ) meaning that the $i$ -th edge connects the vertex $x_i$ from the first part and the vertex $y_i$ from the second part.
The next line contains one integer $q$ ( $1 \le q \le 2 \cdot 10^5$ ) — the number of queries you have to process.
The next $q$ lines contain the queries in the format introduced in the statement.
Additional constraints on the input:
- at any moment, the graph won't contain any multiple edges;
- the queries of type $2$ are only asked if the previous query had type $1$ ;
- there are at most $10$ queries of type $2$ .
输出格式
To answer a query of type $1$ , print one integer — the hash of the optimal coloring.
To answer a query of type $2$ , print one line. It should begin with the integer $k$ — the number of red edges. Then, $k$ distinct integer should follow — the indices of red edges in your coloring, in any order. Each index should correspond to an existing edge, and the hash of the coloring you produce should be equal to the hash you printed as the answer to the previous query.
If there are multiple answers to a query, you may print any of them.
To answer a query of type $2$ , print one line. It should begin with the integer $k$ — the number of red edges. Then, $k$ distinct integer should follow — the indices of red edges in your coloring, in any order. Each index should correspond to an existing edge, and the hash of the coloring you produce should be equal to the hash you printed as the answer to the previous query.
If there are multiple answers to a query, you may print any of them.
输入输出样例
输入 #1
3 4 2 1 2 3 4 10 1 1 3 1 2 3 2 1 3 3 2 1 2 4 2 1 2 1 1 1 1 2
输出 #1
8 8 1 3 40 2 3 5 104 3 5 6 3 104 360 4 5 6 3 8
暂无题解
C++ 编辑器
输入
输出
可保存默认模板;新题优先使用已保存模板。
当前快捷键仅展示,暂不支持修改。
- 撤销
Ctrl / ⌘ + Z - 重做
Ctrl / ⌘ + Y - 查找
Ctrl / ⌘ + F - 全选
Ctrl / ⌘ + A - 复制
Ctrl / ⌘ + C - 剪切
Ctrl / ⌘ + X - 粘贴
Ctrl / ⌘ + V - 自动排版
工具栏排版按钮 - 草稿保存
编辑时自动保存到本机
历史
提交记录
状态说明时间源码
AI
作答助手
你好,我是作答助手。可以问思路、复杂度、样例含义或代码报错原因;不会直接给出完整 AC 代码。
确定要清空代码吗?
提交通过
评测结果:Accepted