hdu 1543 Paint the Wall

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1543


题目大意:给墙壁上色(把某一片正方形涂色),上过色的地方还可以再上色,但是颜色会变为最新刷的颜色,求刷完之后每种颜色的面积和在墙上能看到几个颜色.

题目思路:
不知道谁把这题归到线段树= =,果断的二维区间更新去想,但是想不到外层树怎么更新,本来准备试试矩形树(没用过),结果结果...暴力了.
把x和y离散化,之后暴力更新和暴力求面积.
矩形树的代码也贴了.
不过小数据看不出矩形树的优势的,不过矩形树超过了1000,数组也很难开.

代码(暴力):

#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;

#define ull unsigned __int64
#define ll __int64
//#define ull unsigned long long
//#define ll long long
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define middle (l+r)>>1
#define MOD 1000000007
#define esp (1e-4)
const int INF=0x3F3F3F3F;
const double DINF=10001.00;
//const double pi=acos(-1.0);
const int N=110;
int n,m;
int mtx[N<<1][N<<1],X[N<<1],Y[N<<1],xc,yc,cnt[N];
struct node{
	int x1,y1,x2,y2,c;
	void write(){
		scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&c);
		if(x1>x2) swap(x1,x2);
		if(y1>y2) swap(y1,y2);
	}
}a[N];

int bs(int key,int size,int A[]){
	int l=0,r=size-1,mid;
	while(l<=r){
		mid=middle;
		if(key>A[mid]) l=mid+1;
		else if(key<A[mid]) r=mid-1;
		else return mid;
	}return -1;
}

int main(){
	//freopen("1.in","r",stdin);
	//freopen("1.out","w",stdout);
	int i,j,k,w,h,x1,y1,x2,y2;
	int T,cas=0;//scanf("%d",&T);for(cas=1;cas<=T;cas++)
	while(~scanf("%d%d",&w,&h)){
		if(w==h && !w) break;
		scanf("%d",&n);
		for(i=m=0;i<n;i++){
			a[i].write();
			X[m]=a[i].x1,Y[m]=a[i].y1,m++;
			X[m]=a[i].x2,Y[m]=a[i].y2,m++;
		}
		sort(X,X+m);sort(Y,Y+m);
		for(i=xc=1;i<m;i++) if(X[i]!=X[i-1]) X[xc++]=X[i];
		for(i=yc=1;i<m;i++) if(Y[i]!=Y[i-1]) Y[yc++]=Y[i];
		memset(mtx,0,sizeof(mtx));
		for(i=0;i<n;i++){
			x1=bs(a[i].x1,xc,X),x2=bs(a[i].x2,xc,X);
			y1=bs(a[i].y1,yc,Y),y2=bs(a[i].y2,yc,Y);
			for(j=x1;j<x2;j++)
				for(k=y1;k<y2;k++)
					mtx[j][k]=a[i].c;
		}
		memset(cnt,0,sizeof(cnt));
		for(i=0;i<xc;i++){
			for(j=0;j<yc;j++) if(mtx[i][j]){
				cnt[mtx[i][j]]+=(X[i+1]-X[i])*(Y[j+1]-Y[j]);
			}
		}
		if(cas) puts("");
		printf("Case %d:\n",++cas);
		for(i=1,n=0;i<=100;i++)if(cnt[i]){
			n++;
			printf("%d %d\n",i,cnt[i]);
		}
		if(n==1) printf("There is %d color left on the wall.\n",n);
		else printf("There are %d colors left on the wall.\n",n);
	}
	return 0;
}

(2)矩形树代码

#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;

#define ull unsigned __int64
#define ll __int64
//#define ull unsigned long long
//#define ll long long
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define middle (l+r)>>1
#define MOD 1000000007
#define esp (1e-4)
const int INF=0x3F3F3F3F;
const double DINF=10001.00;
//const double pi=acos(-1.0);
const int N=110;
int n,m;
int X[N<<1],Y[N<<1],xc,yc;
int cov[(N*N)<<6],vis[N],cnt;

struct node{
    int xl,xr,yl,yr;
    int xmid(){return (xl+xr)>>1;}
    int ymid(){return (yl+yr)>>1;}
};

node New(int xl,int xr,int yl,int yr){
    node r;
    r.xl=xl,r.xr=xr;
    r.yl=yl,r.yr=yr;
    return r;
}

struct rct{
    int x1,y1,x2,y2,c;
    void write(){
        scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&c);
        if(x1>x2) swap(x1,x2);
        if(y1>y2) swap(y1,y2);
    }
}a[N];

int bs(int key,int size,int A[]){
    int l=0,r=size-1,mid;
    while(l<=r){
        mid=middle;
        if(key>A[mid]) l=mid+1;
        else if(key<A[mid]) r=mid-1;
        else return mid;
    }return -1;
}

void PushDown(int rt){
    if(cov[rt]!=-1){
        cov[(rt<<2)-2]=cov[(rt<<2)-1]=cov[rt];
        cov[rt<<2]=cov[rt<<2|1]=cov[rt];
        cov[rt]=-1;
    }
}

void Update(node p,int rt,node P,int c){
    if((P.xl<=p.xl&&p.xr<=P.xr) && (P.yl<=p.yl&&p.yr<=P.yr)){
        cov[rt]=c;
        return;
    }
    PushDown(rt);
    int xmid=p.xmid(),ymid=p.ymid();
    if(P.xl<=xmid){
        if(P.yl<=ymid) Update(New(p.xl,xmid,p.yl,ymid),(rt<<2)-2,P,c);
        if(ymid<P.yr) Update(New(p.xl,xmid,ymid+1,p.yr),(rt<<2)-1,P,c);
    }
    if(xmid<P.xr){
        if(P.yl<=ymid) Update(New(xmid+1,p.xr,p.yl,ymid),rt<<2,P,c);
        if(ymid<P.yr) Update(New(xmid+1,p.xr,ymid+1,p.yr),rt<<2|1,P,c);
    }
}

void Query(node p,int rt){
    if(cov[rt]!=-1){
        vis[cov[rt]]+=(X[p.xr+1]-X[p.xl])*(Y[p.yr+1]-Y[p.yl]);
        return;
    }
    if(p.xl==p.xr && p.yl==p.yr) return;
    PushDown(rt);
    int xmid=p.xmid(),ymid=p.ymid();
    Query(New(p.xl,xmid,p.yl,ymid),(rt<<2)-2);
    if(ymid<p.yr) Query(New(p.xl,xmid,ymid+1,p.yr),(rt<<2)-1);
    if(xmid<p.xr){
        Query(New(xmid+1,p.xr,p.yl,ymid),rt<<2);
        if(ymid<p.yr) Query(New(xmid+1,p.xr,ymid+1,p.yr),rt<<2|1);
    }
}

int main(){
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    int i,j,k,r,c,x1,y1,x2,y2;
    int T,cas=0;//scanf("%d",&T);for(cas=1;cas<=T;cas++)
    while(~scanf("%d%d",&r,&c)){
        if(r==c && !r) break;
        scanf("%d",&n);
        for(i=m=0;i<n;i++){
            a[i].write();
            X[m]=a[i].x1,Y[m]=a[i].y1,m++;
            X[m]=a[i].x2,Y[m]=a[i].y2,m++;
        }
        sort(X,X+m);sort(Y,Y+m);
        for(i=xc=1;i<m;i++) if(X[i]!=X[i-1]) X[xc++]=X[i];
        for(i=yc=1;i<m;i++) if(Y[i]!=Y[i-1]) Y[yc++]=Y[i];
        memset(cov,-1,sizeof(cov));
        for(i=0;i<n;i++){
            x1=bs(a[i].x1,xc,X),x2=bs(a[i].x2,xc,X)-1;
            y1=bs(a[i].y1,yc,Y),y2=bs(a[i].y2,yc,Y)-1;
            Update(New(0,xc-2,0,yc-2),1,New(x1,x2,y1,y2),a[i].c);
        }
        memset(vis,0,sizeof(vis));
        Query(New(0,xc-2,0,yc-2),1);
        if(cas) puts("");
        printf("Case %d:\n",++cas);
        for(cnt=0,i=1;i<=100;i++) if(vis[i]){
            cnt++;
            printf("%d %d\n",i,vis[i]);
        }
        if(cnt==1) puts("There is 1 color left on the wall.");
        else printf("There are %d colors left on the wall.\n",cnt);
    }
    return 0;
}