Petrozavodsk Winter-2018. Carnegie Mellon U Contest. A. Mines 线段树建图+强连通分量

Problem A. Mines
Input file: standard input
Output file: standard output
Time limit: 10 seconds

There are N mines on the number line. Mine i is at position pi and has an explosion radius ri. It initially
costs ci to detonate. If mine i is detonated, an explosion occurs on interval [pi - ri; pi + ri] and all mines
in that interval (inclusive of the endpoints) are detonated for free, setting off a chain reaction. You need
to process Q operations of the form (m; c): Change the cost of detonating mine m to c. Output the
minimum cost required to detonate all mines after each change. Note that each change is permanent.

Input

The first line contains integers N and Q (1 ≤ N; Q ≤ 200 000). The next N lines contain information on the mines. The i-th of these lines contains integers pi, ri and ci (1 ≤ pi; ri; ci ≤ 109). The next Q lines each contains space separated integers m and c (1 ≤ m ≤ N, 1 ≤ c ≤ 109).

Output
Output Q lines. The i-th line should contain the minimum cost required to detonate all mines after the
i-th operation

Example

standard input	standard output
4 2 1 1 1 6 3 10 8 2 5 10 2 3 1 1 4 11	4 6

 

 

 

n个矿井分布在一维坐标轴上，分别有不同的爆炸半径和爆炸成本。花费Ci引爆其中一个会引起连锁反应，现在Q次操作，每次改变一个矿井的花费，要求对于每次操作输出最少花费多少引爆所有矿井。

 

建图很容易想，但是边数太多。我们可以把点先按位置排序，再用线段树建图，树上每个点代表一段连续区间所有的点。这样，边数被压缩到nlogn的级别。接着就好求了，只要把图缩点之后，找到DAG当中入度为0的点，在这些点代表的强连通分量当中取一个花费最小的点就好。至于Q次操作，可以对这些强连通分量当中的所有点以及花费用set维护。

时间复杂度O(nlogn)

 

双向dfs求强连通分量的过程：

1.以任意一个点为源点，进行dfs，并将记录经过点的时间戳，时间戳逐渐增加。 
2.进行dfs后，将图中的边的方向反向。寻找时间戳最小的点为源点（就是上面源点）进行dfs。这时，它所能达到的点集就是一个连通分量。并记录搜索过的点 
3.在没有搜索过的点中以时间戳最小的点为源点，继续dfs，搜索结果同上 
4.不断重复3，直到所有点都搜索过。 

 

顺便吐槽一下CSDN,新编辑器居然不能改字体大小了，很多功能找不到，越来越烂了，是逼我转Markdown吗

 

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