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POJ 2528 Mayor's posters(离散+线段树)

时间:2014-08-04 20:47:37      阅读:369      评论:0      收藏:0      [点我收藏+]

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题目大意:往一面墙上贴与墙等高的海报,n次贴完后,求可以看见的海报总数(看见一部分也算)

思路:明显的区间维护,用线段树,不过裸的线段树超时超空间,可以把坐标离散,得到不超过200000个有效点,每个点都表示一个小区间(a[i]~a[i+1]这一段),然后就可以轻松地解决了。不过题目有个坑,给定的右坐标其实不是实际坐标,还要+1,区间总数等于有效点数 -1,更新的时候把查找到的右端点-1再代入,具体见代码:

bubuko.com,布布扣
  1 /**********************************************
  2 ***    Problem:
  3 ***    Author:        JKL
  4 ***    University:    CSUST
  5 ***    Team:          __Dream
  6 ***    Email:          1451108308@QQ.COM
  7 ***    My Blog:        http://www.cnblogs.com/jklongint/
  8 ***********************************************/
  9 //===================================================
 10 #include <iostream>
 11 #include <fstream>
 12 #include <sstream>
 13 #include <iomanip>
 14 #include <cstdio>
 15 #include <cstdlib>
 16 #include <cmath>
 17 #include <cassert>
 18 #include <numeric>
 19 #include <ctime>
 20 #include <algorithm>
 21 #include <cstring>
 22 #include <string>
 23 #include <vector>
 24 #include <queue>
 25 #include <map>
 26 #include <stack>
 27 #include <list>
 28 #include <set>
 29 #include <bitset>
 30 #include <deque>
 31 using namespace std;
 32 //---------------------------------------------------
 33 #define mem(a,b) memset(a,b,sizeof(a))
 34 #define GO cout<<"HelloWorld!"<<endl
 35 #define Case(x) cout<<"Case "<<x<<":"
 36 #define foru(i,n) for(int i=1; i <= n; i++)
 37 #define ford(i,n) for(int i = n; i >= 1; i--)
 38  #define fin freopen("input.txt","r",stdin);
 39  #define fout freopen("output.txt","w",stdout)
 40 #define lson  l, m, rt << 1
 41 #define rson  m + 1, r, rt << 1 | 1
 42 
 43 #define sqr(a)  ((a)*(a))
 44 #define abs(a) ((a>0)?(a):-(a))
 45 #define pii pair<int,int>
 46 
 47 #define fmax(a,b) max(a,b)
 48 #define fmin(a,b) min(a,b)
 49 #define fmax3(a,b,c)  (fmax(a,fmax(a,b)))
 50 #define fmin3(a,b,c)  (fmin(a,fmin(a,b)))
 51 
 52 #define sfi(x) scanf("%d",&x)
 53 #define sfL(x) scanf("%I64d",&x)
 54 #define sfc(x) scanf("%c",&x)
 55 #define sfd(x) scanf("%lf",&x)
 56 #define sfs(x) scanf("%s",x)
 57 #define sfii(a,b) scanf("%d%d",&a,&b)
 58 #define sfLL(a,b) scanf("%I64d%I64d",&a,&b)
 59 #define sfcc(a,b) scanf("%c%c",&a,&b)
 60 #define sfdd(a,b) scanf("%lf%lf",&a,&b)
 61 #define sfss(a,b) scanf("%s%s",a,b)
 62 
 63 #define pfi(x) printf("%d",x)
 64 #define pfL(x) printf("%I64d",x)
 65 #define pfs(x) printf("%s",x)
 66 #define pfd(x) printf("%lf",x)
 67 #define pfc(x) print("%c",x)
 68 #define newLine pfs("\n")
 69 #define space pfs(" ")
 70 
 71 //--------------------------------------------------------
 72 typedef long long LL;
 73 typedef unsigned long long ULL;
 74 typedef __int64 __LL;
 75 typedef unsigned __int64 __ULL;
 76 
 77 typedef vector<int> vi;
 78 typedef vector<LL> vL;
 79 typedef vector<string> vs;
 80 typedef set<int> si;
 81 typedef map<int,int> mii;
 82 typedef map<LL,LL> mLL;
 83 typedef map<string,int> msi;
 84 typedef map<char,int> mci;
 85 //--------------------------------------------------------
 86 const int dx[4]={1,-1,0,0};
 87 const int dy[4]={0,0,1,-1};
 88  const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 89  const int N6=1000006;
 90  const int N5=100006;
 91  const int N4=10006;
 92  const int N3=1006;
 93  const int N2=106;
 94  const int N=20009;
 95  const int MOD=1000000007;
 96  const LL LMAX=0x7fffffffffffffff;
 97  const LL IMAX=0x3fffffff;
 98  const double PI=3.14159265359;
 99 //--------------------------------------------------------
100 template< class T > T gcd(T a, T b) { return (b != 0 ? gcd<T>(b, a%b) : a); }
101 template< class T > T lcm(T a, T b) { return (a / gcd<T>(a, b) * b); }
102 
103 //------------------------------------------------------------
104 struct TreeNode{
105     int  mark;
106 };
107 //=================================================================
108 TreeNode tree[N << 2];
109 int T, n, cnt, flag[N], x[N], y[N], a[N], b[N];
110 void PushUP(int rt)
111 {
112     int tl = tree[rt << 1].mark, tr = tree[rt << 1 | 1].mark;
113     if(tl != tr || tl == -1) tree[rt].mark = -1;
114     else tree[rt].mark = tl;
115 }
116 void PushDOWN(int rt)
117 {
118     int rl = rt << 1, rr = rt << 1 | 1;
119     if(tree[rt].mark > 0){
120         tree[rl].mark = tree[rt].mark;
121         tree[rr].mark = tree[rt].mark;
122     }
123 }
124 void build(int l, int r, int rt)
125 {
126     tree[rt].mark = 0;
127     if(l == r)return;
128     int m = (l + r) >> 1;
129     build(lson);
130     build(rson);
131 }
132 void update(int L, int R, int p, int l, int r, int rt)
133 {
134     if(L <= l && R >= r){
135         tree[rt].mark = p;
136         return;
137     }
138     int m = (l + r) >> 1;
139     PushDOWN(rt);
140     if(L <= m) update(L, R, p, lson);
141     if(R > m) update(L, R, p, rson);
142     PushUP(rt);
143 }
144 int query(int p, int l, int r, int rt)
145 {
146     if(l == r){
147         return tree[rt].mark;
148     }
149     int m = (l + r) >> 1;
150     PushDOWN(rt);
151     if(p <= m)return query(p, lson);
152     else return query(p, rson);
153 }
154 void discrete()
155 {
156     cnt = 0;
157     foru(i, n){
158         a[++cnt] = x[i];
159         a[++cnt] = y[i];
160     }
161     sort(a + 1, a + 1 + cnt);
162     int cc = 0;
163     foru(i, cnt){
164         if(a[i] != a[i - 1] || i == 1)b[++cc] = a[i];
165     }
166     cnt = cc;
167 }
168 int get(int x)
169 {
170     int l = 1, r = cnt;
171     while(l < r){
172         int m = (l + r) >> 1;
173         if(x < b[m]) r = m - 1;
174         if(x == b[m]) return m;
175         if(x > b[m]) l = m + 1;
176     }
177     return l;
178 }
179 int main()
180 {
181     //fin;//fout;//freopen("input.txt","r",stdin);
182    cin>>T;
183    while(T--){
184         cin>>n;
185         foru(i, n)sfii(x[i], y[i]),y[i]++;
186         discrete();
187         build(1, cnt - 1, 1);
188         foru(i, n){
189             int l = get(x[i]), r = get(y[i]) - 1;
190             if(r >= l)update(l, r, i, 1, cnt - 1, 1);
191         }
192         mem(flag, 0);
193         int ans = 0;
194         foru(i, cnt - 1){
195             int c = query(i, 1, cnt - 1, 1);
196             if(!flag[c] && c > 0){
197                 flag[c] = 1;
198                 ans++;
199             }
200         }
201         cout<<ans<<endl;
202    }
203     return 0;
204 }
View Code

 

POJ 2528 Mayor's posters(离散+线段树),布布扣,bubuko.com

POJ 2528 Mayor's posters(离散+线段树)

标签:style   blog   http   color   os   io   for   art   

原文地址:http://www.cnblogs.com/jklongint/p/3890710.html

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