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HDU 4052 Adding New Machine（矩形面积并）

2024-07-25 14:23:24   220人阅读

Adding New Machine


Problem Description
Incredible Crazily Progressing Company (ICPC) suffered a lot with the low speed of procedure. After investigation, they found that the bottleneck was at Absolutely Crowded Manufactory (ACM). In oder to accelerate the procedure, they bought a new machine for ACM. But a new problem comes, how to place the new machine into ACM? 

ACM is a rectangular factor and can be divided into W * H cells. There are N retangular old machines in ACM and the new machine can not occupy any cell where there is old machines. The new machine needs M consecutive cells. Consecutive cells means some adjacent cells in a line. You are asked to calculate the number of ways to choose the place for the new machine. 
 

Input
There are multiple test cases (no more than 50). The first line of each test case contains 4 integers W, H, N, M (1 ≤ W, H ≤ 107, 0 ≤ N ≤ 50000, 1 ≤ M ≤ 1000), indicating the width and the length of the room, the number of old machines and the size of the new machine. Then N lines follow, each of which contains 4 integers Xi1, Yi1, Xi2 and Yi2 (1 ≤ Xi1 ≤ Xi2 ≤ W, 1 ≤ Yi1 ≤ Yi2 ≤ H), indicating the coordinates of the i-th old machine. It is guarantees that no cell is occupied by two old machines. 
 

Output
Output the number of ways to choose the cells to place the new machine in one line. 
 

Sample Input
3 3 1 2
2 2 2 2
3 3 1 3
2 2 2 2
2 3 2 2
1 1 1 1
2 3 2 3
 

Sample Output
8
4
3
 

Author
GUAN, Yao
 

Source
2011 Asia Dalian Regional Contest
 

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题目大意：

     w*h的格子，现在有n个矩形上已经摆放了东西，现在你要放一个东西长度为m，问你有多少种方法？


解题思路：

    枚举没用的情况，如果东西横着放

   （1）如果碰到障碍物，也就是如果 某个东西占着Xi1,Yi1,Xi2,Yi2，那么max(xi1+1-m,0) , x2 y1y2这片矩形区域就不能放东西。

   （2）还有就是被墙当着，也就是max(0,w+1-m),w，0，h这片矩形区域就不能放东西。

   如果竖着放，也是差不多的。

    坑点：如果m==1，也就是横着和竖着是一样的情况，只需考虑一种，我就是这被坑了，弹了20遍。


解题代码：

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

typedef long long ll;

const int maxn=51000;
struct node{
    int l,r,pos,c;
    node(int l0=0,int r0=0,int pos0=0,int c0=0){
        l=l0,r=r0,pos=pos0;c=c0;
    }
    friend bool operator < (node a,node b){
        return a.pos<b.pos;
    }
};

struct rec{
    int x1,y1,x2,y2;
}d[maxn];


vector <node> v;
int w,h,m,n;
vector <int> c;
map <int,int> mp;

struct Tree{
   int l,r,cover,len;
}tree[8*maxn];

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

void pushup(int k){
    if(tree[k].cover>0) tree[k].len=c[tree[k].r]-c[tree[k].l];
    else if(tree[k].l+1==tree[k].r) tree[k].len=0;
    else tree[k].len=tree[k<<1].len+tree[k<<1|1].len;
}

void insert(int l,int r,int k,int c0){
    if(l<=tree[k].l && tree[k].r<=r){
        tree[k].cover+=c0;
    }
    else{
        int mid=(tree[k].l+tree[k].r)>>1;
        if(r<=mid) insert(l,r,k<<1,c0);
        else if(l>=mid) insert(l,r,k<<1|1,c0);
        else{
            insert(l,mid,k<<1,c0);
            insert(mid,r,k<<1|1,c0);
        }
    }
    pushup(k);
}

ll getans(){
    if(v.size()<=0) return (ll)w*(ll)h;
    c.clear();
    mp.clear();
    sort(v.begin(),v.end());
    for(int i=0;i<v.size();i++){
        mp[v[i].l]=i;
        mp[v[i].r]=i;
    }
    for(map <int,int>::iterator it=mp.begin();it!=mp.end();it++){
        it->second=c.size();
        c.push_back(it->first);
    }
    ll ret=0;
    build(0,c.size()-1,1);
    insert(mp[v[0].l],mp[v[0].r],1,v[0].c);
    for(int i=1;i<v.size();i++){
        ret+=(ll)(v[i].pos-v[i-1].pos)*(ll)tree[1].len;
        insert(mp[v[i].l],mp[v[i].r],1,v[i].c);
    }
    return (ll)w*(ll)h-ret;
}

void solve(){
    v.clear();
    for(int i=0;i<n;i++){
        v.push_back(node(max(d[i].x1+1-m,0),d[i].x2,d[i].y1,1));
        v.push_back(node(max(d[i].x1+1-m,0),d[i].x2,d[i].y2,-1));
    }
    if(m>1){
        v.push_back(node(max(0,w+1-m),w,0,1));
        v.push_back(node(max(0,w+1-m),w,h,-1));
    }

    ll ans=getans();

    if(m==1){
        printf("%I64d\n",ans);
        return;
    }

    v.clear();
    for(int i=0;i<n;i++){
        v.push_back(node(max(d[i].y1+1-m,0),d[i].y2,d[i].x1,1));
        v.push_back(node(max(d[i].y1+1-m,0),d[i].y2,d[i].x2,-1));
    }
    if(m>1){
        v.push_back(node(max(0,h+1-m),h,0,1));
        v.push_back(node(max(0,h+1-m),h,w,-1));
    }
    ans+=getans();
    printf("%I64d\n",ans);
}

int main(){
    while(scanf("%d%d%d%d",&w,&h,&n,&m)!=EOF){
        for(int i=0;i<n;i++){
            scanf("%d%d%d%d",&d[i].x1,&d[i].y1,&d[i].x2,&d[i].y2);
            d[i].x1--;d[i].y1--;
        }
        solve();
    }
    return 0;
}





HDU 4052 Adding New Machine（矩形面积并）


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