Can of Worms 【迭代/线段树】

题意：一条直线上有n个炸弹，给出每个炸弹的爆炸半径，可以引爆另一个炸弹爆炸。问：每个炸弹爆炸后，最多有几个炸弹一起爆炸？

 

迭代，用线段树更新。

 

  1 #include <cstdio>
  2 #include <algorithm>
  3 #include <iostream>
  4 #define ll long long
  5 #define lson l, m, rt<<1
  6 #define rson m+1, r, rt<<1|1
  7 #define X first
  8 #define Y second
  9 #define mp make_pair
 10 #define pii pair<int, int>
 11 #define gg puts("gg");
 12 using namespace std;
 13 const int N = 1e5+5;
 14 struct p{
 15     int n, pos, ra;
 16     p(){}
 17     p(int pos, int ra):pos(pos), ra(ra){}
 18 };
 19 bool cmppos(p A, p B){
 20     return A.pos < B.pos;
 21 }
 22 
 23 p pp[N];
 24 //after sort
 25 int pos[N];
 26 int lpos[N], rpos[N], now;
 27 
 28 struct Node{
 29     int l, r;
 30     Node(){}
 31     Node(int l, int r):l(l), r(r){}
 32 };
 33 Node T[N<<2];
 34 void pushup(int rt){
 35     T[rt].l = min(T[rt<<1].l, T[rt<<1|1].l);
 36     T[rt].r = max(T[rt<<1].r, T[rt<<1|1].r);
 37 }
 38 void gao(Node& x, Node y){
 39     if(x.l > y.l) x.l = y.l;
 40     if(x.r < y.r) x.r = y.r;
 41 }
 42 void update(int x, Node y, int l, int r, int rt){
 43     if(x == l&&x == r){
 44         T[rt] = y;
 45         return ;
 46     }
 47     int m = l+r >> 1;
 48     if(x <= m) update(x, y, lson);
 49     else update(x, y, rson);
 50     pushup(rt);
 51 }
 52 void build(int l, int r, int rt){
 53     if(l == r){
 54         T[rt].l = lpos[now];
 55         T[rt].r = rpos[now];
 56         now++;
 57         return ;
 58     }
 59     int m = l+r >> 1;
 60     build(lson);
 61     build(rson);
 62     pushup(rt);
 63 }
 64 void query(Node& ans, int L, int R, int l, int r, int rt){
 65     if(L <= l&& r <= R){
 66         gao(ans, T[rt]);
 67         return ;
 68     }
 69     int m = l+r >> 1;
 70     if(L <= m) query(ans, L, R, lson);
 71     if(m < R) query(ans, L, R, rson);
 72 }
 73 int ans[N];
 74 void debug(int i){
 75     printf("%d: lpos:%d, rpos:%d\n", i, lpos[i], rpos[i]);
 76 }
 77 
 78 int main(){
 79     int n;
 80     while(scanf("%d", &n), n){
 81         for(int i = 1; i <= n; i++){
 82             pp[i].n = i;
 83             scanf("%d%d", &pp[i].pos, &pp[i].ra);
 84             pos[i] = pp[i].pos;
 85         }
 86         sort(pos+1, pos+n+1);
 87         sort(pp+1, pp+n+1, cmppos);
 88 
 89         for(int i = 1; i <= n; i++)
 90             lpos[i] = pp[i].pos-pp[i].ra, rpos[i] = pp[i].pos+pp[i].ra;
 91 
 92         now = 1;
 93         build(1, n, 1);
 94 
 95         bool tag = true;
 96         while(tag){
 97             tag = false;
 98             for(int i = 1; i <= n; i++){
 99                 int lcan = lower_bound(pos+1, pos+n+1, lpos[i])-pos, rcan = upper_bound(pos+1, pos+n+1, rpos[i])-pos-1;
100                 ans[ pp[i].n ] = rcan-lcan+1;
101                 Node ret = Node(2e9, -2e9);
102                 query(ret, lcan, rcan, 1, n, 1);
103                 if(lpos[i] > ret.l||rpos[i] < ret.r) {
104                     tag = true;
105                     if(lpos[i] > ret.l) lpos[i] = ret.l;
106                     if(rpos[i] < ret.r) rpos[i] = ret.r;
107                     update(i, ret, 1, n, 1);
108                 }
109             }
110         }
111         for(int i = 1; i <= n; i++)
112             printf("%d%c", ans[i], " \n"[i == n]);
113     }
114     return 0;
115 }
View Code

 

posted @ 2016-10-25 14:39  我在地狱  阅读(271)  评论(0)  编辑  收藏  举报