线段树的简单实现

线段树的简单实现：＃i nclude <stdio.h>＃i nclude <alloc.h>#define MaxSize 1000typedef char ElemType;typedef struct node{int i,j;int cover;struct node *lchild,*rchild;} ITree;//建立线段树ITree *CreateTree(int a,int b){ITree *r;if(b<a) return NULL;r=(ITree *)malloc(sizeof(ITree));r->i=a;r->j=b;r->cover=0;if(b-a>1){r->lchild=CreateTree(a,(a+b)/2);r->rchild=CreateTree((a+b)/2,b);}else r->lchild=r->rchild=NULL;return r;}//线段树的插入void InsTree(ITree *r,int a,int b){int mid;if(a<=r->i&&r->j<=b) r->cover++;else{mid=(r->i+r->j)/2;if(b<=mid) InsTree(r->lchild,a,b);else if(mid<=a) InsTree(r->rchild,a,b);else{InsTree(r->lchild,a,mid);InsTree(r->rchild,mid,b);}}}//线段树的删除void DelTree(ITree *r,int a,int b){int mid;if(a<=r->i&&r->j<=b) r->cover--;else{mid=(r->i+r->j)/2;if(b<=mid) DelTree(r->lchild,a,b);else if(mid<=a) DelTree(r->rchild,a,b);else{DelTree(r->lchild,a,mid);DelTree(r->rchild,mid,b);}}}//计算线段树的测度int Count(ITree *r){if(r->cover>0) return r->j-r->i;else if(r->j-r->i==1) return 0;else return Count(r->lchild)+Count(r->rchild);}//主函数int main(){ITree *r;int a,b,a1,b1,a2,b2,a3,b3,n,i;printf("请输入线段树的区间端点：");while(scanf("%d%d",&a,&b)!=EOF){r=CreateTree(a,b);printf("请输入插入3条线段的区间端点：");scanf("%d%d%d%d%d%d",&a1,&b1,&a2,&b2,&a3,&b3);InsTree(r,a1,b1);InsTree(r,a2,b2);InsTree(r,a3,b3);printf("插入3条线段后线段树的测度为%d\n",Count(r));printf("请输入删除2条线段的区间端点：");scanf("%d%d%d%d",&a1,&b1,&a2,&b2);DelTree(r,a1,b1);DelTree(r,a2,b2);printf("删除2条线段后线段树的测度为%d\n",Count(r));printf("\n请输入线段树的区间端点：");}return 0; }pku 2528(离散化+线段树)2010-03-16 16:42#include<iostream> #include<algorithm>using namespace std;struct node{int left,right;int count;}tree[160020];int poster[10003][2];int flag[20003];int hash[10000003];//hash函数。

int sum;int cmp(const void *a,const void *b){return *(int *)a-*(int *)b;}void Build(int k,int i,int j)//利用完全二叉树建树。

{tree[k].left=i;tree[k].right=j;tree[k].count=0;if(i!=j){int mid=(i+j)/2;Build(2*k,i,mid);Build(2*k+1,mid+1,j);}return ;}void Insert(int i,int color,int from,int to){if(tree[i].left==from&&tree[i].right==to){tree[i].count=color;return ;}if(tree[i].count==color)return ;if(tree[i].count!=-1&&tree[i].count!=color)//这个if语句很重要，就是他，搞了很长时间。

{tree[2*i].count=tree[i].count;tree[2*i+1].count=tree[i].count;tree[i].count=-1;}int mid=(tree[i].left+tree[i].right)>>1;if(to<=mid){Insert(i*2,color,from,to);}else if(from>mid)Insert(1+2*i,color,from,to);else{Insert(i*2,color,from,mid);Insert(i*2+1,color,mid+1,to);}return ;}void query(int i){if(tree[i].count>=0){if(flag[tree[i].count]==0){sum++;}flag[tree[i].count]++;return ;}query(i*2);query(i*2+1);return ;}int main(){int i,j,k;int t,n;scanf("%d",&t);while(t--){memset(flag,0,sizeof(flag));scanf("%d",&n);j=0;k=1;for(i=1;i<=n;i++){scanf("%d%d",&poster[i][0],&poster[i][1]);flag[j++]=poster[i][0];flag[j++]=poster[i][1];}qsort(flag,j,sizeof(flag[0]),cmp);//离散化处理。

hash[flag[0]]=1;for(i=1;i<j;i++){if(flag[i]!=flag[i-1])k++;hash[flag[i]]=k;}Build(1,1,k);//建立空的线段树memset(flag,0,sizeof(flag));for(i=1;i<=n;i++){Insert(1,i,hash[poster[i][0]],hash[poster[i][1]]);//插入}sum=0;query(1);//计算可以看到总的海报数量printf("%d\n",sum);}return 0;}pku 2777(线段树)2010-03-16 18:49#include<iostream> using namespace std;const int hmax=100005;int flag[100];int sum=0;struct node{int left,right;int color;}board[4*hmax];void Build(int k,int from,int to){board[k].left=from;board[k].right=to;board[k].color=1;if(from!=to){int mid=(from+to)>>1;Build(2*k,from,mid);Build(2*k+1,mid+1,to);}return ;}void update(int i,int from,int to,int color){if(board[i].left==from&&board[i].right==to){board[i].color=color;return ;}if(board[i].color==color)return ;if(board[i].color!=color&&board[i].color){board[2*i].color=board[i].color;board[2*i+1].color=board[i].color;board[i].color=0;}int mid=(board[i].left+board[i].right)>>1;if(to<=mid){update(i*2,from,to,color);}else if(from>mid)update(i*2+1,from,to,color);else{update(i*2,from,mid,color);update(i*2+1,mid+1,to,color);}return ;}void query(int i,int from,int to){if(board[i].color){if(!flag[board[i].color]){sum++;}flag[board[i].color]++;return ;}int mid=(board[i].left+board[i].right)>>1; if(to<=mid){query(i*2,from,to);}else if(from>mid)query(i*2+1,from,to);else{query(i*2,from,mid);query(i*2+1,mid+1,to);}return ;}int main(){int i,a,b,c,k;int L,T,O;char s1[3];while(scanf("%d%d%d",&L,&T,&O)!=EOF) {Build(1,1,L);for(i=0;i<O;i++){scanf("%s",s1);if(s1[0]=='C'){scanf("%d%d%d",&a,&b,&c);if(a>b){k=a;a=b;b=k;}update(1,a,b,c);}else{scanf("%d%d",&a,&b);if(a>b){k=a;a=b;b=k;}sum=0;memset(flag,0,sizeof(flag));query(1,a,b);printf("%d\n",sum);}}}return 0;}pku 2777(线段树)2010-03-16 18:49#include<iostream> using namespace std;const int hmax=100005;int flag[100];int sum=0;struct node{int left,right;int color;}board[4*hmax];void Build(int k,int from,int to){board[k].left=from;board[k].right=to;board[k].color=1;if(from!=to){int mid=(from+to)>>1;Build(2*k,from,mid);Build(2*k+1,mid+1,to);}return ;}void update(int i,int from,int to,int color){if(board[i].left==from&&board[i].right==to){board[i].color=color;return ;}if(board[i].color==color)return ;if(board[i].color!=color&&board[i].color){board[2*i].color=board[i].color;board[2*i+1].color=board[i].color;board[i].color=0;}int mid=(board[i].left+board[i].right)>>1;if(to<=mid){update(i*2,from,to,color);}else if(from>mid)update(i*2+1,from,to,color);else{update(i*2,from,mid,color);update(i*2+1,mid+1,to,color);}return ;}void query(int i,int from,int to){if(board[i].color){if(!flag[board[i].color]){sum++;}flag[board[i].color]++;return ;}int mid=(board[i].left+board[i].right)>>1; if(to<=mid){query(i*2,from,to);}else if(from>mid)query(i*2+1,from,to);else{query(i*2,from,mid);query(i*2+1,mid+1,to);}return ;}int main(){int i,a,b,c,k;int L,T,O;char s1[3];while(scanf("%d%d%d",&L,&T,&O)!=EOF) {Build(1,1,L);for(i=0;i<O;i++){scanf("%s",s1);if(s1[0]=='C'){scanf("%d%d%d",&a,&b,&c);if(a>b){k=a;a=b;b=k;}update(1,a,b,c);}else{scanf("%d%d",&a,&b);if(a>b){k=a;a=b;b=k;}sum=0;memset(flag,0,sizeof(flag));query(1,a,b);printf("%d\n",sum);}}}return 0;}pku 1177(线段树)2010-03-30 11:48强烈推荐先看看国家集训队的论文---陈宏的论文#include<iostream>#include<algorithm>using namespace std;const int hmax=40006;struct seg_tree{int left,right;int count;int cl,cr;int segment;int seg;}node[2*hmax];struct rect{int start,end;int x;int tag;}line[hmax];int hash[hmax],lisan[hmax];;int Y[hmax],n;int cmp(rect a,rect b){return (a.x<b.x)? 1:0;}void Build(int i,int from,int to){node[i].left=from;node[i].right=to;node[i].count=0;node[i].cl=node[i].cr=0;node[i].segment=0;node[i].seg=0;if(to-from>1){int mid=(1+node[i].left+node[i].right)>>1;Build(2*i,from,mid);Build(2*i+1,mid,to);}return;}void update(int i,int from,int to){if(node[i].count>0){node[i].segment=lisan[node[i].right]-lisan[node[i].left];node[i].seg=1;node[i].cl=node[i].cr=1;}else if(node[i].right-node[i].left>1){node[i].segment=(node[2*i].segment+node[2*i+1].segment);node[i].cl=node[2*i].cl;node[i].cr=node[2*i+1].cr;node[i].seg=node[2*i].seg+node[2*i+1].seg;if(node[2*i].cr&&node[2*i+1].cl)node[i].seg--;}else{node[i].segment=0;node[i].cl=node[i].cr=0;node[i].seg=0;}}void Insert(int i,int from,int to,int tag){if(from<=node[i].left&&to>=node[i].right)node[i].count+=tag;else{int mid=(1+node[i].left+node[i].right)>>1;if(from<mid)Insert(2*i,from,to,tag);if(to>mid)Insert(2*i+1,from,to,tag);}update(i,from,to);}int main(){int i,j,k;int x1,x2,y1,y2;while(EOF!=scanf("%d",&n)){for(i=j=0;j<n;j++){scanf("%d%d%d%d",&x1,&y1,&x2,&y2);line[i].start=line[i+1].start=y1;line[i].end=line[i+1].end=y2;line[i].x=x1;line[i+1].x=x2;line[i].tag=1;line[i+1].tag=-1;Y[i]=y1;Y[i+1]=y2;i+=2;}sort(Y,Y+2*n);sort(line,line+2*n,cmp);/*for(i=0;i<2*n;i++)cout<<Y[i]<<endl;cout<<endl;for(i=0;i<2*n;i++)cout<<line[i].x<<" "<<line[i].start<<" "<<line[i].end<<" "<<line[i].tag<<endl;cout<<endl;*/lisan[1]=Y[0];hash[Y[0]+10000]=1;k=1;for(i=1;i<2*n;i++){if(Y[i]==Y[i-1])continue;k++;lisan[k]=Y[i];hash[Y[i]+10000]=k;}Build(1,1,k);int from,to;int pre=0, len=0;int sum=0;line[2*n].x=line[2*n-1].x;for(i=0;i<2*n;i++){from=hash[line[i].start+10000];to=hash[line[i].end+10000];Insert(1,from,to,line[i].tag);len=node[1].segment;sum+=2*node[1].seg*(line[i+1].x-line[i].x);if(len>=pre)sum+=len-pre;elsesum+=pre-len;pre=len;}printf("%d\n",sum);}return 0;}/*-6-4481010141516202225-15 0 10 1-5 8 25 10 -6 4 12 15 22 15 0 10 -110 15 22 -115 -4 14 116 -6 4 -120 8 25 -1 24 -4 14 -1 30 10 20 1 34 0 16 1 36 10 20 -1 40 0 16 -1POJ 2182 Lost Cows 我的第一道线段树题目！DescriptionN (2 <= N <= 8,000) cows have unique brands in the range 1..N. In a spectacular display of poor judgment, they visited the neighborhood 'watering hole' and drank a few too many beers before dinner. When it was time to line up for their evening meal, they did not line up in the required ascending numerical order of their brands.Regrettably, FJ does not have a way to sort them. Furthermore, he's not very good at observing problems. Instead of writing down each cow's brand, he determined a rather silly statistic: For each cow in line, he knows the number of cows that precede that cow in line that do, in fact, have smaller brands than that cow.Given this data, tell FJ the exact ordering of the cows.Input* Line 1: A single integer, N* Lines 2..N: These N-1 lines describe the number of cows that precede a given cow in line and have brands smaller than that cow. Of course, no cows precede the first cow in line, so she is not listed. Line 2 of the input describes the number of preceding cows whose brands are smaller than the cow in slot #2; line 3 describes the number of preceding cows whose brands are smaller than the cow in slot #3; and so on.Output* Lines 1..N: Each of the N lines of output tells the brand of a cow in line. Line #1 of the output tells the brand of the first cow in line; line 2 tells the brand of the second cow; and so on. Sample Input5121Sample Output24531解题思路：对于一个给定序列，从后面往前面看是前面有几个数字是比他小的，0代表前面没有数字比他小，所以肯定是1，把1去掉。