UVALive 7141 BombX(离散化+线段树)(2014 Asia Shanghai Regional Contest)

题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&category=648&page=show_problem&problem=5153

In an infinite chess board, some pawns are placed on some cells.
You have a rectangular bomb that is W width and H height.
The bomb’s orientation is fixed, you can’t rotate it. The bomb
can only be placed on an entirely unoccupied area. The bomb
explodes both horizontally and vertically, killing all pawns that
are in the cross shape (see picture on the right).
Your mission is to choose the placement of the bomb, and
maximize the number of bombed pawns.
The picture corresponds to the first test case in the sample.
Input
The first line of the input gives the number of test cases, T. T
test cases follow. Each test case starts with a line containing N,
W, H, indicating number of pawns, width of the bomb, height of
the bomb, respectively.
N lines follow. Each line contains 2 integers: x, y, indicating there is a pawn on cell (x, y). No two
pawns are in the same cell.
Output
For each test case, output one line containing ‘Case #x: y’, where x is the test case number (stating
from 1) and y is the maximum number of bombed pawns.

 

题目大意:无穷大的地图上有n个兵,现给出一个W*H的炸弹(不可以旋转),一次能炸死它所在的行和列的所有小兵,不能放在小兵的上面。问:一次最多能干掉多少个小兵?

思路:最初的想法就是,枚举所有可以放炸弹的位置,看看哪个位置可以炸的小兵最多。如果用a[x, y]来表示炸弹放在以(x, y)为左下角的矩阵里,能炸多少小兵(不能放的时候为0)。那么显然结果就是a[x, y]的最大值。(作用:对拍)

然而这并不能AC。我们现在改成只考虑a数组的一个维度,让x一步一步往右移动,修改a数组。当我们在x位置的时候,用a[y]来表示,炸弹放在以(x, y)为左下角的矩阵里,能炸多少小兵。假设我们不考虑能不能放的问题,当x不断往右移动(不断+1)的时候,a数组并不会发生任何变化。我们可以另开一个数组cnt[],当我们遇到一个点(x0, y0)的时候,给cnt[y0-h+1..y0]都加上一个1,表示有一个点导致这个区间不能放入炸弹(同理当某个点离开区间[x..x+w-1]的时候,对应的区间都减去1)。那么我们在移动x的时候,要求的就是max{a[y] | cnt[y]=0}。

然而这还是不能AC。注意到在x往右移动的过程中,若没有点进出区间[x..x+w-1]的时候,cnt[]数组是不会发生变化的,这时,离散化x坐标的作用就发挥出来了。同时,y坐标也是可以离散化的,因为每次更改的区间都是可以提前预知的。

此时复杂度已经变成了O(n^2),当然还是不够的。注意到每次修改cnt[]数组都是区间修改,这里使用线段树优化,复杂度变为O(nlogn),可以AC。

此题细节多多,大家多想想。

PS:范围写的是w, h≥0,于是我加了一个assert,目前数据是没有w, h=0的情况的。如果rejudge RE了我就知道怎么回事啦!

 

代码(0.386S):

  1 #include <cstdio>
  2 #include <algorithm>
  3 #include <iostream>
  4 #include <cstring>
  5 #include <vector>
  6 #include <tuple>
  7 #include <cassert>
  8 using namespace std;
  9 typedef long long LL;
 10 
 11 const int MAXN = 100010;
 12 const int MAXL = 200010;
 13 const int MAXT = 400010;
 14 
 15 struct Node {
 16     int x, y;
 17     void read() {
 18         scanf("%d%d", &x, &y);
 19     }
 20 } p[MAXN];
 21 
 22 int val[MAXL];
 23 
 24 vector<int> ytmp;
 25 
 26 int T, n, w, h;
 27 
 28 void inithash() {
 29     ytmp.clear();
 30     for(int i = 0; i < n; ++i) {
 31         ytmp.push_back(p[i].y - h + 1);
 32         ytmp.push_back(p[i].y + 1);
 33     }
 34     sort(ytmp.begin(), ytmp.end());
 35     ytmp.erase(unique(ytmp.begin(), ytmp.end()), ytmp.end());
 36 }
 37 
 38 int yhash(int y) {
 39     return lower_bound(ytmp.begin(), ytmp.end(), y) - ytmp.begin();
 40 }
 41 
 42 void initval() {
 43     memset(val, 0, ytmp.size() * sizeof(int));
 44     for(int i = 0; i < n; ++i) {
 45         val[yhash(p[i].y - h + 1)]++;
 46         val[yhash(p[i].y + 1)]--;
 47     }
 48     partial_sum(val, val + ytmp.size(), val);
 49 }
 50 
 51 int root(int l, int r) {
 52     return (l + r) | (l != r);
 53 }
 54 #define mid ((l + r) >> 1)
 55 #define rt root(l, r)
 56 #define ll root(l, mid)
 57 #define rr root(mid + 1, r)
 58 int ban[MAXT], maxt[MAXT];
 59 
 60 void update(int l, int r) {
 61     if(!ban[rt]) maxt[rt] = (l == r ? val[l] : max(maxt[ll], maxt[rr]));
 62     else maxt[rt] = 0;
 63 }
 64 
 65 void build(int l, int r) {
 66     if(l == r) {
 67         maxt[rt] = val[l];
 68     } else {
 69         build(l, mid);
 70         build(mid + 1, r);
 71         update(l, r);
 72     }
 73 }
 74 
 75 void modify_ban(int l, int r, int a, int b, int val) {
 76     if(a <= l && r <= b) {
 77         ban[rt] += val;
 78     } else {
 79         if(a <= mid) modify_ban(l, mid, a, b, val);
 80         if(mid < b) modify_ban(mid + 1, r, a, b, val);
 81     }
 82     update(l, r);
 83 }
 84 
 85 int query() {
 86     return maxt[root(0, ytmp.size() - 1)];
 87 }
 88 
 89 vector<int> allx;
 90 vector<tuple<int, int, int> > xtmp;
 91 #define gpos(t) get<0>(t)
 92 #define gval(t) get<1>(t)
 93 #define gy(t) get<2>(t)
 94 
 95 int solve() {
 96     xtmp.clear();
 97     for(int i = 0; i < n; ++i) {
 98         xtmp.push_back(make_tuple(p[i].x - w + 1, 1, p[i].y));
 99         xtmp.push_back(make_tuple(p[i].x + 1, -1, p[i].y));
100     }
101     sort(xtmp.begin(), xtmp.end());
102 
103     allx.clear();
104     for(int i = 0; i < n; ++i) {
105         allx.push_back(p[i].x - w + 1);
106         allx.push_back(p[i].x + 1);
107     }
108     sort(allx.begin(), allx.end());
109     allx.erase(unique(allx.begin(), allx.end()), allx.end());
110 
111     int res = query(), cnt = 0;
112     size_t i = 0;
113     for(int x : allx) {
114         while(i < xtmp.size() && gpos(xtmp[i]) <= x) {
115             auto t = xtmp[i++];
116             modify_ban(0, ytmp.size() - 1, yhash(gy(t) - h + 1), yhash(gy(t) + 1) - 1, gval(t));
117             cnt += gval(t);
118         }
119         res = max(res, query() + cnt);
120     }
121     return res;
122 }
123 
124 int main() {
125     scanf("%d", &T);
126     for(int t = 1; t <= T; ++t) {
127         scanf("%d%d%d", &n, &w, &h);
128         assert(w > 0 && h > 0);
129         for(int i = 0; i < n; ++i)
130             p[i].read();
131         inithash();
132         initval();
133         build(0, ytmp.size() - 1);
134         printf("Case #%d: %d\n", t, solve());
135     }
136 }
View Code

 

posted @ 2015-05-15 21:35  Oyking  阅读(375)  评论(0编辑  收藏