2016 UESTC Training for Data Structures (1)

来源：互联网发布：稻谷出糙率计算数据编辑：程序博客网时间：2024/04/30 18:57

用了1mol多时间+借鉴(chao)了1mol多题解后终于差不多a完了(话说直接把题目copy过来让我们做真的好?)
A 线段树模板不用多说了吧

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>#include<stack>#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define sdrwr 1using namespace std;long long n,m,t,a,b;long long sum[400005];void pushup(long long rt){    sum[rt]=max(sum[rt<<1],sum[rt<<1|1]);}void update(long long rt,long long l,long long r,long long q,long long x){    if (l==r)    {        sum[rt]=sum[rt]+x;        return;    }    long long mid=(l+r)>>1;    if (q<=mid)update(lson,q,x);    else update(rson,q,x);    pushup(rt);}long long query(long long rt,long long l,long long r,long long x,long long y){    if (l>=x&&r<=y)    {        return sum[rt];    }    long long mid=(l+r)>>1;    long long ans=0;    if (x<=mid)ans=max(ans,query(lson,x,y));    if (y>mid)ans=max(ans,query(rson,x,y));    return ans;}int main(){      scanf("%lld%lld",&n,&m);    for (long long i=1;i<=m;i++)    {        scanf("%lld%lld%lld",&t,&a,&b);        if (t==1) {update(1,1,n,a,b);}        if (t==2) {printf("%lld\n",query(1,1,n,a,b));}    }    return 0;}

B 卿学姐与基本法
离散化的线段树,但我在网上看到了一种神奇的算法

#include <cstdio>#include <cmath>#include <iostream>#include <cstring>#include <algorithm>#include <map>using namespace std;map <int,int> q;map <int,int> ::iterator w;int t,l,r,x,y,haha;int query(int x,int y){    int l=x,r=y,ans=y-x+1;    for (w=q.begin();w!=q.end();w++)    {        int a=w->first,b=w->second;        if (a>r) return ans;        else if (b<l) continue;        else if (a<l&&b>r)         {            return ans-=r-l+1;        }           else         {         if(a<=l && b<=r)                ans-=b-l+1,l=b+1;            else if(a>l && b<=r)                ans-=b-a+1,l=b+1;            else                return ans-=r-a+1;        }       }    return ans;}int main(){    int n,t;    scanf("%d%d",&n,&t);    for (int i=1;i<=t;i++)    {        scanf("%d%d%d",&x,&l,&r);        if (x==1)        {            if(q[l]<r) q[l]=r;        }        else printf("%d\n",query(l,r));    }}

C 卿学姐与诡异村庄
并查集,开两倍内存,如果是一个人另外一个人说是好人,就说明是同类,反之不是.所以如果是一就find(x,y);和find(x+n,y+n);否则find(x,y+n)和find(x+n,y);如果有矛盾就懵逼,否则就行.

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>#include<stack>using namespace std;int fa[200005],x,t,n,sdf;int find(int x){    if (fa[x]==x) return fa[x];    else return fa[x]=find(fa[x]);}int main(){    scanf("%d",&n);    for (int i=1;i<=2*n;i++)    fa[i]=i;    for (int i=1;i<=n;i++)    {        scanf("%d%d",&x,&t);        if (t==1)         {            int x2=find(x);            int y2=find(i);            if (x2!=y2)            {                fa[x2]=y2;            }            x2=find(x+n);            y2=find(i+n);            if (x2!=y2)            {                fa[x2]=y2;            }        }        else        {            int x2=find(x+n);            int y2=find(i);            if (x2!=y2)            {                fa[x2]=y2;            }            x2=find(x);            y2=find(i+n);            if (x2!=y2)            {                fa[x2]=y2;            }        }    }    for (int i=1;i<=n;i++)    {        if (find(i)==find(i+n))        {            printf("One face meng bi");            return 0;        }    }    printf("Time to show my power");    return 0;}

D 卿学姐与魔法
优先队列,排序,再把所有(a[i]+b[1],1)压进去就Ok,弹出时把b的下一个压进去即可,水题一枚

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>#include<stack>using namespace std;int n,a[100005],b[100005],d[100005],ans,cnt=1,min1,min2;typedef pair<int,int >pii;int main(){    scanf("%d",&n);    for (int i=1;i<=n;i++)    {        scanf("%d",&a[i]);    }    sort(a+1,a+1+n);    for (int i=1;i<=n;i++)    {        scanf("%d",&b[i]);    }    sort(b+1,b+1+n);    priority_queue<pii, vector<pii>, greater<pii> > q;    for (int i=1;i<=n;i++)    {        q.push(make_pair(a[i]+b[1],1));    }    int t=1;     for (int i=1;i<=n;i++)    {        pii u=q.top();        q.pop();                q.push(make_pair(u.first-b[u.second]+b[u.second+1],u.second+1));        printf("%d\n",u.first);    }    return 0;}

E 卿学姐与城堡的墙
离散化树状数组,蛤蛤蛤,

#include <cstdio>  #include <cstdlib>  #include <algorithm>  using namespace std;  #define ll long long  #define maxn 200005  struct Line  {      ll y1, y2;      bool operator < (Line x)const      {          return y1 < x.y1;      }  }line[maxn];  int N;  ll u, v, data[maxn];  ll tree[maxn];  void update(int i, ll val)  {      while (i <= N)      {          tree[i] += val;          i += i&(-i);      }  }  ll query(int i)  {      ll sum = 0;      while (i)      {          sum += tree[i];          i -= i&(-i);      }      return sum;  }  int main()  {       scanf("%d%lld%lld", &N, &u, &v);      ll a, b;      for (int i = 0; i < N; ++i)      {          scanf("%lld%lld", &a, &b);          line[i].y1 = a*u + b, line[i].y2 = a*v + b;          data[i] = line[i].y2;      }      sort(line, line + N);      sort(data, data + N);      int num = unique(data, data + N) - data;      for (int i = 0; i < N; ++i)          line[i].y2 = lower_bound(data, data + num, line[i].y2) - data + 1;      ll ans = 0;      int last = 0;      for (int i = 0; i < N; ++i)      {          if (i > 0 && line[i].y1 == line[i - 1].y1)          {              //++ans;              ans += i - last;          }          else          {              ans += i - query(line[i].y2 - 1);              last = i;          }          update(line[i].y2, 1);      }      printf("%lld\n", ans);      return 0;  }

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