Tsinghua dsa mooc pa1

第一题Range

关键：二分查找，查找不大于一个数的最大下标。

 #include <cstdlib>
 #include <cstdio> 
 4 int compare (const void * a, const void * b){
     return ( *(int*)a - *(int*)b );
 }
 int bisearch(int *array,int low, int high,int aim){
     int mid;
     while(low<=high){
         mid = (low+high)>>;
         if(aim<array[mid])
             high=mid-;
         else if(aim>array[mid])
             low=mid+;
         else
             return mid;
     }
 //    printf("%d\n",high);
     return high;
 }
 int main(int argc, char* argv[]){
 //    freopen("in.txt", "r", stdin);
     int pnum,qtime;
     scanf("%d %d",&pnum,&qtime);
     int *points = new int[pnum];
     for(int i=; i<pnum; i++)
         scanf("%d",&points[i]);
     qsort(points,pnum,sizeof(int),compare);
     int lower,higher;//range
     int left,right;
     for(int i=; i<qtime; i++){
         scanf("%d %d",&lower,&higher);
         left=bisearch(points,,pnum-,lower);
         right=bisearch(points,,pnum-,higher);
         if(right<){//notice
             printf("0\n");
             continue;
         }
         if(left<){//notice
             printf("%d\n",right+);
             continue;
         }
         if(points[left]==lower){
             printf("%d\n",right-left+);
         }else
             printf("%d\n",right-left);
     }
     delete [] points;
     return ;
 }

　　当使用cin/cout时候，Exceed Time Limit；此外，我还比较了使用静态分配数组和动态堆分配数组的运行时间差异，发现几乎相同。

第二题祖玛，自己尝试写了一个简单的双向链表，能进行插入和删除。20个案例，最后一个超时。此外，这个题也是C/C++混杂起来使用，好尴尬。

 #include <cstdio>
 #include <string>
 #include <iostream>
 using namespace std;
 class Node{
 public:
     Node* prev;
     Node* next;
     char content;
     Node():prev(NULL),next(NULL),content(){}
     Node(char c):prev(NULL),next(NULL),content(c){}
 };
 class Dolist{
 public:
     Node head;
     int len;
     Dolist(string in);
     ~Dolist();
     Node* insert(int pos, char ch);
     void del(Node *inpos);
     void print();
 };
 Dolist::~Dolist(){
     Node *fir=head.next;
     while(fir){
         Node *next=fir->next;
         delete fir;
         fir=next;
     }
 }
 Node* Dolist::insert(int pos, char ch){
     Node *tmp = new Node(ch);
     Node* cur=&head;
     //pos<=len
     if(pos==len){//insert in the end
         while(cur->next){
             cur=cur->next;
         }
         cur->next=tmp;
         tmp->prev=cur;
     }else{
         for(int i=;i<pos;i++){
             cur=cur->next;
         }
         Node* nex=cur->next;
         cur->next=tmp;
         tmp->prev=cur;
         tmp->next=nex;
         nex->prev=tmp;
     }
     len++;
     return tmp;
 }
 void Dolist::del(Node *inpos){
     if(inpos==&head)
         return;
     int dellen=;//num of same nodes
     Node *fir=inpos,*las=inpos;//pointers between same nodes,from beginning to end
     while(fir->content==fir->prev->content){//search left
         dellen++;
         fir=fir->prev;
     }
     while(las->next){//search right
         if(las->content==las->next->content){
             dellen++;
             las=las->next;
         }else
             break;
     }
     if(dellen>=){//if num is above to 3,delete
         Node *tmp=fir->prev;
         fir->prev->next=las->next;
         if(las->next)
             las->next->prev=fir->prev;
         Node *tmp1=fir->next;
         while(tmp1!=las){
             delete fir;
             fir=tmp1;
             tmp1=fir->next;
         }
         delete fir;
         delete las;
         len -= dellen;
         del(tmp);//iteration

     }
 }
 void Dolist::print(){
     if(head.next==NULL)
         printf("-\n");
     else{
         Node *tmp=head.next;
         while(tmp){
             putchar(tmp->content);
             tmp=tmp->next;
         }
         printf("\n");
     }
 }
 Dolist::Dolist(string in){
     len=;
     int siz=in.size();
     for(int i=; i<siz; i++){
         insert(i,in[i]);
     }
 }
 int main(int argc, char* argv[]){
     //freopen("in.txt","r",stdin);
     string init;
     getline(cin,init);
     Dolist zuma(init);//initial
     int time;
     scanf("%d",&time);
     int pos;
     char ch;
     while(time--){
         scanf("%d %c",&pos,&ch);
         Node *inpos = zuma.insert(pos,ch);
         zuma.del(inpos);
         zuma.print();
     }
     return ;
 }

第三题灯塔，先对x坐标使用快速排序，排序后对y坐标进行归并排序，在归并排序的同时求出顺序对的数目。20个案例通过19个，最后一个超时。

 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 using namespace std;
 typedef struct tower{
     int x;
     int y;
 }Tower;
 int compare(const void *a,const void *b){
     return ((*(Tower*)a).x - (*(Tower*)b).x);
 }
 long long Merge(int *array, int m, int mid, int n){
     long long tmp=;
     int *copy=new int[n-m+];
     int i,j,k=;
     for(i=m,j=mid+;i<=mid&&j<=n;){
         if(array[i]<array[j]){
             copy[k]=array[i];
             tmp += (n-j+);//这是最关键的，在进行merge时，如果array[i]<array[j]，那么从j到n的数都大于array[i]
             i++;
         }else{
             copy[k]=array[j];
             j++;
         }
         k++;
     }
     if(i<=mid){
         for(int ii=i; ii<=mid; ii++){
             copy[k]=array[ii];
             k++;
         }
     }else if(j<=n){
         for(int ii=j; ii<=n; ii++){
             copy[k]=array[ii];
             k++;
         }
     }
     memcpy(array+m,copy,(n-m+)*sizeof(int));
     delete [] copy;
     return tmp;
 }
 long long MergeSort(int *array, int m, int n){
     if(m==n) return ;
     int mid = m+((n-m)>>);
     long long num1 = MergeSort(array,m,mid);
     long long num2 = MergeSort(array,mid+,n);
     long long num3 = Merge(array,m,mid,n);
     return num1+num2+num3;
 }

 int main(int argc,char *argv[]){
     //freopen("in.txt","r",stdin);
     int num;
     scanf("%d",&num);
     Tower *all = new Tower[num];
     for(int i=; i<num; i++){
         scanf("%d %d",&(all[i].x),&(all[i].y));
     }
     qsort(all,num,sizeof(Tower),compare);
     int *ally = new int[num];//取出所有的y,通过归并排序求出顺序对
     for(int i=; i<num; i++)
         ally[i]=all[i].y;
     long long out = MergeSort(ally,,num-);
     printf("%lld\n",out);
     delete [] all;
     delete [] ally;
     return ;
 }