POJ2528+线段树
见代码。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 /* 2 线段树+Lazy 3 题意:有一面墙,被等分为1QW份,一份的宽度为一个单位宽度。 4 现在往墙上贴N张海报,每张海报的宽度是任意的,但是必定是单位宽度的整数倍,且<=1QW。 5 后贴的海报若与先贴的海报有交集,后贴的海报必定会全部或局部覆盖先贴的海报。 6 现在给出每张海报所贴的位置(左端位置和右端位置),问张贴完N张海报后,还能看见多少张海报? 7 (PS:看见一部分也算看到。) 8 */ 9 #include<stdio.h> 10 #include<string.h> 11 #include<stdlib.h> 12 #include<algorithm> 13 #include<iostream> 14 #include<queue> 15 #include<map> 16 #include<stack> 17 #include<set> 18 #include<math.h> 19 using namespace std; 20 typedef long long int64; 21 //typedef __int64 int64; 22 typedef pair<int64,int64> PII; 23 #define L(x) (x<<1) 24 #define R(x) (x<<1|1) 25 #define MP(a,b) make_pair((a),(b)) 26 const int maxn = 20005; 27 const int inf = 0x7fffffff; 28 const double pi=acos(-1.0); 29 const double eps = 1e-8; 30 31 struct Node{ 32 int L,R,color; 33 }tree[ maxn<<2 ]; 34 struct NODE{ 35 int L,R; 36 bool operator<(const NODE &tmp ) const{ 37 return R<tmp.R; 38 } 39 }a[ maxn ]; 40 int myhash[ maxn ]; 41 int ANS; 42 int vis[ maxn ]; 43 44 int lisanhua( int k ){ 45 sort( myhash,myhash+k ); 46 int cnt = unique( myhash,myhash+k ) - myhash; 47 return cnt; 48 } 49 50 void build( int L,int R,int n ){ 51 if( L==R ){ 52 tree[ n ].L = L; 53 tree[ n ].R = R; 54 tree[ n ].color = 0; 55 return ; 56 } 57 tree[ n ].L = L; 58 tree[ n ].R = R; 59 tree[ n ].color = 0; 60 int mid = (L+R)/2; 61 build( L,mid,L(n) ); 62 build( mid+1,R,R(n) ); 63 } 64 65 void PushDown( int n,int new_color ){ 66 if( tree[ n ].color!=0&&tree[ n ].color!=new_color ){ 67 tree[ L(n) ].color = tree[ R(n) ].color = tree[ n ].color; 68 tree[ n ].color = 0; 69 } 70 } 71 72 void update( int x,int y,int new_color,int L,int R,int n ){ 73 if( x==L&&y==R ){ 74 tree[ n ].color = new_color; 75 return ; 76 } 77 PushDown( n,new_color ); 78 int mid = (L+R)/2; 79 if( mid>=y ) update( x,y,new_color,L,mid,L(n) ); 80 else if( mid<x ) update( x,y,new_color,mid+1,R,R(n) ); 81 else { 82 update( x,mid,new_color,L,mid,L(n) ); 83 update( mid+1,y,new_color,mid+1,R,R(n) ); 84 } 85 } 86 87 void query( int n ){ 88 if( tree[n].color ){ 89 if( vis[tree[n].color]==0 ){ 90 vis[tree[n].color] = 1; 91 ANS++; 92 } 93 return ; 94 } 95 //int mid = (tree[n].L+tree[n].R)/2; 96 query( L(n) ); 97 query( R(n) ); 98 } 99 100 int main(){ 101 int T; 102 scanf("%d",&T); 103 while( T-- ){ 104 int n; 105 scanf("%d",&n); 106 int k = 0; 107 for( int i=1;i<=n;i++ ){ 108 scanf("%d%d",&a[i].L,&a[i].R); 109 myhash[ k++ ] = a[i].L; 110 myhash[ k++ ] = a[i].R; 111 } 112 int cnt = lisanhua( k ); 113 build( 1,cnt,1 ); 114 //printf("build\n"); 115 for( int i=1;i<=n;i++ ){ 116 int x = lower_bound( myhash,myhash+cnt,a[i].L )-myhash; 117 int y = lower_bound( myhash,myhash+cnt,a[i].R )-myhash; 118 x++,y++; 119 //printf("i = %d\n",i); 120 //printf("x = %d, y=%d\n",x,y); 121 update( x,y,i,1,cnt,1 ); 122 } 123 ANS = 0; 124 memset( vis,0,sizeof( vis ) ); 125 //printf("query\n"); 126 query( 1 ); 127 printf("%d\n",ANS); 128 } 129 return 0; 130 }
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