hdu 4022 Bombing(STL)

原创于 2018-04-08 19:35:37 发布 · 476 阅读

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本文介绍了一个典型的轰炸路径计算问题，通过使用C++实现的算法解决了如何高效地计算轰炸次数及目标摧毁数量的问题。采用map和multiset数据结构进行坐标离散化处理，确保了算法效率。

Bombing

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 4834 Accepted Submission(s): 1813

Problem Description

It’s a cruel war which killed millions of people and ruined series of cities. In order to stop it, let’s bomb the opponent’s base.
It seems not to be a hard work in circumstances of street battles, however, you’ll be encountered a much more difficult instance: recounting exploits of the military. In the bombing action, the commander will dispatch a group of bombers with weapons having the huge destructive power to destroy all the targets in a line. Thanks to the outstanding work of our spy, the positions of all opponents’ bases had been detected and marked on the map, consequently, the bombing plan will be sent to you.
Specifically, the map is expressed as a 2D-plane with some positions of enemy’s bases marked on. The bombers are dispatched orderly and each of them will bomb a vertical or horizontal line on the map. Then your commanded wants you to report that how many bases will be destroyed by each bomber. Notice that a ruined base will not be taken into account when calculating the exploits of later bombers.

Input

Multiple test cases and each test cases starts with two non-negative integer N (N<=100,000) and M (M<=100,000) denoting the number of target bases and the number of scheduled bombers respectively. In the following N line, there is a pair of integers x and y separated by single space indicating the coordinate of position of each opponent’s base. The following M lines describe the bombers, each of them contains two integers c and d where c is 0 or 1 and d is an integer with absolute value no more than 10 ⁹, if c = 0, then this bomber will bomb the line x = d, otherwise y = d. The input will end when N = M = 0 and the number of test cases is no more than 50.

Output

For each test case, output M lines, the ith line contains a single integer denoting the number of bases that were destroyed by the corresponding bomber in the input. Output a blank line after each test case.

Sample Input

  3 2 1 2 1 3 2 3 0 1 1 3 0 0
 

Sample Output

2 1

Source

The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest

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lcy

题意：

在二维坐标轴上有N个坐标，有M次操作，每次给出d,x当d=0时，删除x轴上坐标为x的所有坐标并输出个数，d=1时，删除y轴上坐标为x的所有坐标输出个数。

思路：

因为只有1e5个坐标，也就是x和y轴上出现的坐标最多也只有1e5，而数可以达到1e9，需要离散化。

所以用map+multiset来解决，设置两个一个存x轴上的每个坐标对应的y坐标，同理一个存y轴的。

每次更新的时候都需要将两个map来更新。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
using namespace std;

map <int,multiset<int> > mpx;
map <int,multiset<int> > mpy;
multiset<int> :: iterator it;
int main()
{
    int n,m,x,y;
    while(~scanf("%d%d",&n,&m)&&n&&m)
    {
        mpx.clear();
        mpy.clear();
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&x,&y);
            mpx[x].insert(y);
            mpy[y].insert(x);
        }
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&x,&y);
            if(x==0)
            {
                printf("%d\n",mpx[y].size());
                for(it=mpx[y].begin();it!=mpx[y].end();it++)
                {
                    mpy[*it].erase(mpy[*it].find(y));
                }
                mpx[y].clear();
            }
            else
            {
                printf("%d\n",mpy[y].size());
                for(it=mpy[y].begin();it!=mpy[y].end();it++)
                {
                    mpx[*it].erase(mpx[*it].find(y));
                }
                mpy[y].clear();
            }
        }
        printf("\n");
    }
    return 0;
}