在平面坐标系里给出一些矩形,求面积并.
Atlantis

#include<cstdio>
#include<map>
#include<algorithm>
#include<iostream>
using namespace std;

const int N=106;
int n,m,num=0;

struct P{
    double x,y,z;
    int k;
    bool operator<(const P w) const{
        return x<w.x;
    }
}a[N<<1];

double raw[N<<1];
map<double ,int > val ;

struct Tree{
    int l,r,cnt;
    double len;
}t[N<<3];

void build(int p,int l,int r){
    t[p].l=l;
    t[p].r=r;
    t[p].cnt=0;
    t[p].len=0;
    if(l==r) return ;
    int mid=(l+r)>>1;
    build(p<<1,l,mid);
    build(p<<1|1,mid+1,r);
}

void change(int p,int l,int r,int k){
    if(l<=t[p].l&&r>=t[p].r) t[p].len=((t[p].cnt+=k)>0?raw[t[p].r+1]-raw[t[p].l]:0);
    if(t[p].l==t[p].r) return;
    int mid=(t[p].l+t[p].r)>>1;
    if(l<=mid) change(p<<1,l,r,k);
    if(r>mid) change(p<<1|1,l,r,k);
    t[p].len=(t[p].cnt>0?raw[t[p].r+1]-raw[t[p].l]:t[p<<1].len+t[p<<1|1].len);
}

void  Atlantis(){
    for(int i=1;i<=n;++i){
        int k=i<<1;
        double y,z; //竖直方向坐标
        scanf("%lf %lf %lf %lf",&a[k-1].x,&y,&a[k].x,&z);
        raw[k-1]=a[k-1].y=a[k].y=y;//矩形左边界
        raw[k]=a[k-1].z=a[k].z=z; //矩形右边界
        a[k-1].k=1;
        a[k].k=-1;
    }
    n<<=1;
    //离散化
    sort(raw+1,raw+n+1);
    int m=unique(raw+1,raw+n+1)-(raw+1);
    for(int i=1;i<=m;++i) val[raw[i]]=i;
    sort(a+1,a+n+1);
    build(1,1,m-1);
    double ans=0;
    for(int i=1;i<n;++i){
        int y=val[a[i].y],z=val[a[i].z]-1;  // y:[val[a[i].y], val[a[i].y]+1]  z: [val[a[i].z]-1, val[a[i].z]]
        change(1,y,z,a[i].k); //y~z: [val[a[i].y], val[a[i].z]
        ans+=t[1].len*(a[i+1].x-a[i].x); //t[1].len 整个扫描线覆盖的长度
    }
    printf("Test case #%d\nTotal explored area: %.2f\n\n", ++num, ans);
}

int main() {
    while (cin >> n && n) Atlantis();
    return 0;
}

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int SIZE = 100010;

struct SegmentTree{
    int l,r;
    long long sum,add;
    #define l(x) tree[x].l
    #define r(x) tree[x].r
    #define sum(x) tree[x].sum
    #define add(x) tree[x].add
} tree[SIZE*4];

int a[SIZE],n,m;

void build(int p,int l,int r){
    l(p)=l,r(p)=r;
    if(l==r) { sum(p)=a[l];return; }
    int mid=(l+r)/2;
    build(p*2,l,mid);
    build(p*2+1,mid+1,r);
    sum(p)=sum(p*2)+sum(p*2+1);
}

void spread(int p){
    if(add(p)){
        sum(p*2)+=add(p)*(r(p*2)-l(p*2)+1);
        sum(p*2+1)+=add(p)*(r(p*2+1)-l(p*2+1)+1);
        add(p*2)+=add(p);
        add(p*2+1)+=add(p);
        add(p)=0;
    }
}

void change(int p,int l,int r,int d){
    if(l<=l(p) && r>=r(p)){
        sum(p)+=(long long) d*(r(p)-l(p)+1);
        add(p)+=d;
        return;
    }
    spread(p);
    int mid=(l(p)+r(p))/2;
    if(l<=mid) change(p*2,l,r,d);
    if(r>mid) change(p*2+1,l,r,d);
    sum(p)=sum(p*2)+sum(p*2+1);
}

long long ask(int p,int l,int r){
    if(l<=l(p) && r>=r(p)) return sum(p);
    spread(p);
    int mid=(l(p)+r(p))/2;
    long long val=0;
    if(l<=mid) val+=ask(p*2,l,r);
    if(r>mid) val+=ask(p*2+1,l,r);
    return val;
}

int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i){
        scanf("%d",&a[i]);
    }
    build(1,1,n);
    while(m--){
        char op[2];
        int l,r,d;
        scanf("%s%d%d",op,&l,&r);
        if(op[0]=='C'){
            scanf("%d",&d);
            change(1,l,r,d);
        }
        else printf("%lld\n",ask(1,l,r));
    }
    return 0;
}

RT,线段树单点修改,区间查询.

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#include <utility>
using namespace std;
const int INF = 1e9;
const int maxn = 1e6;

#define SIZE 500000

int a[SIZE+1];
struct SegmentTree {
    int l,r;
    int dat,sum,lmax,rmax;
    SegmentTree() {
        l = r = sum = 0;
        lmax = rmax  =dat= -INF;
    }
} t[SIZE*4];

void build(int p,int l,int r) {
    t[p].l=l,t[p].r=r;
    if(l==r) {
        t[p].dat=t[p].lmax=t[p].rmax=t[p].sum=a[l];
        return;
    }
    int mid=(l+r)/2;
    build(p*2,l,mid);
    build(p*2+1,mid+1,r);
    t[p].dat=max(max(t[p*2].dat,t[p*2+1].dat),t[p*2].rmax+t[p*2+1].lmax);
    t[p].lmax=max(t[p*2].lmax,t[p*2].sum+t[p*2+1].lmax);
    t[p].rmax=max(t[p*2+1].rmax,t[p*2+1].sum+t[p*2].rmax);
    t[p].sum=t[p*2].sum+t[p*2+1].sum;
}

void change(int p,int x,int v) {
    if(t[p].l==t[p].r) {
        t[p].dat=t[p].lmax=t[p].rmax=t[p].sum=v;
        return;
    }
    int mid=(t[p].l+t[p].r)/2;
    if(x<=mid)
        change(p*2,x,v);
    else
        change(p*2+1,x,v);
    t[p].dat=max(max(t[p*2].dat,t[p*2+1].dat),t[p*2].rmax+t[p*2+1].lmax);
    t[p].lmax=max(t[p*2].lmax,t[p*2].sum+t[p*2+1].lmax);
    t[p].rmax=max(t[p*2+1].rmax,t[p*2+1].sum+t[p*2].rmax);
    t[p].sum=t[p*2].sum+t[p*2+1].sum;
}

SegmentTree ask(int p,int l,int r) {
    if(l<=t[p].l&&r>=t[p].r) {
        return t[p];
    }
    SegmentTree res,a,b;
    int mid=(t[p].l+t[p].r)/2;
    if(l<=mid) {
        a=ask(p*2,l,r);
        res.sum+=a.sum;
    }
    if(r>mid) {
        b=ask(p*2+1,l,r);
        res.sum+=b.sum;
    }
    res.lmax=max(a.lmax,a.sum+b.lmax);
    res.rmax=max(b.rmax,b.sum+a.rmax);
    res.dat=max(max(a.dat,b.dat),a.rmax+b.lmax);
    return res;
}

int main() {
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1; i<=n; ++i) {
        scanf("%d",&a[i]);
    }
    build(1,1,n);
    while(m--) {
        int k,x,y;
        scanf("%d%d%d",&k,&x,&y);
        if(k==1) {
            if(x>y)
                swap(x,y);
            printf("%d\n",ask(1,x,y).dat);
        } else {
            change(1,x,y);
        }
    }
    return 0;
}

题目链接:CH4201

对某个点,求前面和后面分别有多少个点的纵坐标是比它大的,两数相乘就是以该点为中间点,能构成的"v"的数量.

"^"同理.

code:

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <string>
#include <vector>

using namespace std;

#define maxn 200005

typedef long long ll;
typedef unsigned long long ull;

int c[maxn];
int a[maxn];
int Right1[maxn];
int Left1[maxn];
int Right2[maxn];
int Left2[maxn];
int n;

int ask(int x){
  int ans=0;
  for(;x;x-=x&-x) ans+=c[x];
  return ans;
}

void add(int x,int y){
  for(;x<=n;x+=x&-x) c[x]+=y;
}

int main() {
  cin>>n;
  for(int i=1;i<=n;++i){
    cin>>a[i];
  }
  for(int i=n;i>0;--i){
      Right1[i]=n-i-ask(a[i]);
      Right2[i]=ask(a[i]-1);
      add(a[i],1);
  }
  memset(c,0,sizeof(c));
  for(int i=1;i<=n;++i){
    Left1[i]=i-1-ask(a[i]);
    Left2[i]=ask(a[i]-1);
    add(a[i],1);
  }
  ll ans1=0,ans2=0;

  for(int i=1;i<=n;++i){
    ans1+=(ll)Left1[i]*Right1[i];
    ans2+=(ll)Left2[i]*Right2[i];
  } 
  cout<<ans1<<" "<<ans2<<endl;
  return 0;
}

题目链接:BZOJ 4195

在数据范围大,但实际数据个数不大的时候,采用离散化避免超出数组空间.

code:

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <string>
#include <vector>

using namespace std;

#define maxn 1000005

typedef long long ll;
typedef unsigned long long ull;

int fa[maxn * 2];

int x1[maxn];
int x2[maxn];
int s[maxn];
int n;
vector<int> b;

void init() {
  b.clear();
  for (int i = 0; i <= 2 * n; ++i) {
    fa[i] = i;
  }
}

int get(int x) {
  if (x == fa[x]) return x;
  return fa[x] = get(fa[x]);
}

void merge(int x, int y) { fa[get(x)] = get(y); }

void discrete() {
  sort(b.begin(), b.end());
  unique(b.begin(), b.end());
}
int query(int x) { return lower_bound(b.begin(), b.end(), x) - b.begin(); }

int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    init();
    for (int k = 1; k <= n; ++k) {
      cin >> x1[k] >> x2[k] >> s[k];
      b.push_back(x1[k]);
      b.push_back(x2[k]);
    }
    discrete();
    for (int i = 1; i <= n; ++i) {
      if (s[i]) {
        int x = query(x1[i]);
        int y = query(x2[i]);
        merge(x, y);
      }
    }
    int flag = 0;
    for (int i = 1; i <= n; ++i) {
      if (!s[i]) {
        int x = query(x1[i]);
        int y = query(x2[i]);
        if (get(x) == get(y)) {
          cout << "NO" << endl;
          flag = 1;
          break;
        }
      }
    }
    if (!flag) {
      cout << "YES" << endl;
    }
  }
  return 0;
}