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codeforces CF903G Yet Another Maxflow Problem 線段樹

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G. Yet Another Maxflow Problem


time limit per test: 4 seconds
memory limit per test: 256 megabytes
input: standard input
output: standard output

 

In this problem you will have to deal with a very special network.
 
The network consists of two parts: part $ A $ and part $ B $ .

Each part consists of n vertices; $ i $ -th vertex of part $ A $ is denoted as $ A_i $ , and $ i $ -th vertex of part $ B $ is denoted as $ B_i $ .
 
For each index $ i (1?≤?i?<?n) $ there is a directed edge from vertex $ A_i $ to vertex $ A_{i?+?1} $ ,
and from $ B_i $ to $ B_{i?+?1} $ , respectively.
Capacities of these edges are given in the input.
Also there might be several directed edges going from part $ A $ to part $ B $ (but never from $ B $ to $ A $ ).
 
You have to calculate the maximum flow value from $ A_1 $ to $ B_n $ in this network.
Capacities of edges connecting $ A_i $ to $ A_{i?+?1} $ might sometimes change,
and you also have to maintain the maximum flow value after these changes.
Apart from that, the network is fixed (there are no changes in part $ B $ ,
no changes of edges going from $ A $ to $ B $ , and no edge insertions or deletions).
 
Take a look at the example and the notes to understand the structure of the network better.
 

Input

The first line contains three integer numbers $ n, m $ and $ q (2?≤?n,?m?≤?2·10^5, 0?≤?q?≤?2·10^5) $
— the number of vertices in each part, the number of edges going from $ A $ to $ B $ and the number of changes, respectively.
 
Then $ n?-?1 $ lines follow, $ i $ -th line contains two integers $ x_i $ and $ y_i $ denoting
that the edge from $ A_i $ to $ A_{i?+?1} $ has capacity $ x_i $ and the edge from $ B_i $ to $ B_{i?+?1} $ has capacity $ y_i (1?≤?x_i,?y_i?≤?10^9) $ .
 
Then $ m $ lines follow, describing the edges from $ A $ to $ B $ .
Each line contains three integers $ x, y $ and $ z $ denoting an edge from $ A_x $ to $ B_y $ with capacity $ z (1?≤?x,?y?≤?n, 1?≤?z?≤?10^9) $ .
There might be multiple edges from $ A_x $ to $ B_y $ .
 
And then $ q $ lines follow, describing a sequence of changes to the network.
$ i $ -th line contains two integers $ v_i $ and $ w_i $ , denoting that the capacity of the edge from $ A_{v_i} $ to $ A_{v_i?+?1} $ is set to $ w_i (1?≤?vi?<?n, 1?≤?wi?≤?10^9) $ .
 

Output

Firstly, print the maximum flow value in the original network.
Then print $ q $ integers, $ i $ -th of them must be equal to the maximum flow value after $ i $ -th change.
 

Examples

input

 4 3 2
 1 2
 3 4
 5 6
 2 2 7
 1 4 8
 4 3 9
 1 100
 2 100

output

 9
 14
 14

 

Note

This is the original network in the example:

技術分享圖片

 

題目翻譯

  • 一張圖分為兩部分,左右都有 $ n $ 個節點,

  • $ A_i \rightarrow A_{i+1} $ 連邊, $ B_i \rightarrow B_{i+1} $ 連邊,容量給出

  • 有 $ m $ 對 $ A_i \rightarrow B_j $ 有邊,容量給出

  • 兩種操作

  • 1.修改某條 $ A_i \rightarrow A_{i+1} $ 的邊的容量

  • 2.詢問從 $ A_1 $ 到 $ B_n $ 的最大流

  • $ 2 \le n,m \le 2 \times 10^5 $ 流量 $ \le 10^9 $

感謝@yybyyb 提供的翻譯
 

思路:

  • 因為最大流的流量等於最小割的容量,所以最大流不好做的時候考慮最小割。

  • 一些顯而易見的結論:

  • 假如在 $ A $ 中,割掉 $ A_x $ 和 $ A_{x+1} $ 之間的有向邊,
    那麽割掉 $ A_y (y>x) $ 與 $ A_{y+1} $ 之間的有向邊是沒有意義的。

  • 假如在 $ B $ 中,割掉 $ B_X $ 和 $ B_{x+1} $ 之間的有向邊,
    那麽割掉 $ B_y (y>x) $ 與 $ B_{y+1} $ 之間的有向邊是沒有意義的。

  • 因此,在 $ A, B $ 中至多有一條邊屬於割集。

  • 假設在 $ A $ 中被割掉的邊是 $ (A_x, A_{x+1}) $ ,在 $ B $ 中被割掉的邊是 $ (B_y, B_{y+1}) $ ,
    那麽任意 $ (A_u, B_v) (u \ge x $ 且 $ v> y) $ 都是屬於這個割集的。

  • 因此,割的大小可以看作三部分:在 $ A $ 中的割集大小,在 $ B $ 中的割集大小,在 $ A, B $ 間的割集大小。

  • 由於後兩部分不會改變(只改 $ A $ 的邊),考慮計算出它們。

  • 考慮在 $ A $ 中從小到大枚舉 $ A_x $ ,表示割掉 $ A_x $ 和 $ A_{x+1} $ 之間的邊, $ x=n $ 表示不存在這條邊。

  • 那麽下面要做的事情是在 $ B $ 中找到一個最優決策點 $ y $ ,使得後兩部分的和最小。

  • 對於每加入一條 $ A, B $ 間的邊,對答案造成的影響是連續的一段,從 $ 1 $ 開始。

  • 所以我們只需要寫一個支持區間加,求 $ [1, n] $ 的最小值的線段樹就好了。

  • 合並兩部分的答案。這個可以用一個可刪堆或者線段樹來維護。

  • 時間復雜度 $ O((n+m+q)\quad log_n ) $

codeforces CF903G Yet Another Maxflow Problem 線段樹