2528 Mayor's posters 解题报告

 

 很久米做线段树的题了，昨天看了一下江离的解题报告。发现一个好东西--离散化，这题里面贴海报的范围很广0~10000000，尝试了几次原始的线段树都mle了。现在的代码可以很轻松的就把它给ac了，这就是离散化的伟大之处...

 

#include<stdio.h>#include<algorithm>#include<memory.h>using namespace std;#define NIL 10000001#define MUL -1#define M 20000struct TNode{int left,right;int no;};TNode pst[10*M];int point[3*M],sz;int map[M][3];//根据下标找出坐标在point数组中的位置int find(int x){int left(0),right(sz);int mid = (left + right) >> 1;while(x != point[mid]){if(point[mid] > x)right = mid;elseleft = mid;mid = (left + right) >> 1;}return mid;}void Build(int s,int t,int step){pst[step].left = s;pst[step].right = t;pst[step].no = 0;if(s == t)return;int mid = (s + t)>> 1;Build(s,mid,step*2);Build(mid+1,t,step*2+1);}void Insert(int s,int t,int step,int num){if(pst[step].left == s && pst[step].right == t){pst[step].no = num;return;}if(pst[step].no != MUL){pst[step*2].no = pst[step].no;pst[step*2 + 1].no = pst[step].no;pst[step].no = MUL;}int mid = (pst[step].left + pst[step].right) >> 1;if(mid >= t){Insert(s,t,step*2,num);return;}if(mid < s){Insert(s,t,step*2 + 1,num);return;}Insert(s,mid,step*2,num);Insert(mid+1,t,step*2+1,num);return;}bool v[M];int res;void Count(int step){if(pst[step].no == MUL){Count(step*2);Count(step*2+1);}else{if( !v[pst[step].no]){res++;v[pst[step].no] = true;}}return;}int main(){int c,n,tmp,l,r,i;scanf("%d",&c);while(c--){sz = 0;res = 0;scanf("%d",&n);memset(v,false,sizeof(v));v[0] = true;for(i = 1;i <= n; i++){scanf("%d%d",&l,&r);point[sz++] = l;point[sz++] = r;//将每幅海报的数据都记录下来map[i][0] = l;map[i][1] = r;map[i][2] = i;}//找出重复出现的坐标，将其删掉for(i = 0; i < sz; i++)for(int j = i + 1; j < sz; j++)if(point[i] != point[j])break;elsepoint[j] = NIL;sort(point,point+sz);while(point[sz-1] == NIL)sz--;//根据坐标建线段树Build(0,sz,1);for(i = 1; i <= n; i++)Insert(find(map[i][0]),find(map[i][1]),1,map[i][2]);Count(1);printf("%d/n",res);}return 0;}