poj 1177 Picture

题目链接：http://poj.org/problem?id=1177

Picture Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 10280   Accepted: 5448 Description A number of rectangular posters, photographs and other pictures of the same shape are pasted on a wall. Their sides are all vertical or horizontal. Each rectangle can be partially or totally covered by the others. The length of the boundary of the union of all rectangles is called the perimeter.  Write a program to calculate the perimeter. An example with 7 rectangles is shown in Figure 1.  The corresponding boundary is the whole set of line segments drawn in Figure 2.  The vertices of all rectangles have integer coordinates.  Input Your program is to read from standard input. The first line contains the number of rectangles pasted on the wall. In each of the subsequent lines, one can find the integer coordinates of the lower left vertex and the upper right vertex of each rectangle. The values of those coordinates are given as ordered pairs consisting of an x-coordinate followed by a y-coordinate.  0 <= number of rectangles < 5000  All coordinates are in the range [-10000,10000] and any existing rectangle has a positive area. Output Your program is to write to standard output. The output must contain a single line with a non-negative integer which corresponds to the perimeter for the input rectangles. Sample Input 7 -15 0 5 10 -5 8 20 25 15 -4 24 14 0 -6 16 4 2 15 10 22 30 10 36 20 34 0 40 16 Sample Output 228 Source IOI 1998

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#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<vector>
#include<bitset>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<cstdlib>
using namespace std;
#define CLR(A) memset(A,0,sizeof(A))
struct Node{
    int st,ed,m,lbd,rbd;
    int sequence_line,cnt;
}ST[40005];
void build(int st,int ed,int v){
    ST[v].st=st;ST[v].ed=ed;
    ST[v].m=ST[v].lbd=ST[v].rbd=0;
    ST[v].sequence_line=ST[v].cnt=0;
    if(ed-st>1){
        int mid=(st+ed)>>1;
        build(st,mid,2*v+1);
        build(mid,ed,2*v+2);
    }
}
inline void UpDate(int v){
    if(ST[v].cnt>0){
        ST[v].m=ST[v].ed-ST[v].st;
        ST[v].lbd=ST[v].rbd=1;
        ST[v].sequence_line=1;
        return;
    }
    if(ST[v].ed-ST[v].st==1){
        ST[v].m=0;
        ST[v].lbd=ST[v].rbd=0;
        ST[v].sequence_line=0;
    }
    else{
        int left=2*v+1,right=2*v+2;
        ST[v].m=ST[left].m+ST[right].m;
        ST[v].sequence_line=ST[left].sequence_line+ST[right].sequence_line-(ST[left].rbd&ST[right].lbd);
        ST[v].lbd=ST[left].lbd;
        ST[v].rbd=ST[right].rbd;
    }
}
void Insert(int s,int e,int v){
    if( s<=ST[v].st && e>=ST[v].ed){
        ST[v].cnt++;
        UpDate(v);
        return;
    }
    int mid=(ST[v].st+ST[v].ed)>>1;
    if(s<mid) Insert(s,e,2*v+1);
    if(e>mid) Insert(s,e,2*v+2);
    UpDate(v);
}
void Delete(int st,int ed,int v){
    if(st<=ST[v].st&&ed>=ST[v].ed){
        ST[v].cnt--;
        UpDate(v);
        return;
    }
    int mid=(ST[v].st+ST[v].ed)>>1;
    if(st<mid) Delete(st,ed,2*v+1);
    if(ed>mid) Delete(st,ed,2*v+2);
    UpDate(v);
}
struct line{
    int x,y1,y2;bool d;
}a[10003];
bool cmp(line t1,line t2){return t1.x<t2.x;}
void cal_c(int n){
    int i,j,k,t2=0,sum=0;
    a[n]=a[n-1];
    for(int i=0;i<n;i++){
        if(a[i].d==1) Insert(a[i].y1,a[i].y2,0);
        else Delete(a[i].y1,a[i].y2,0);
        sum+=ST[0].sequence_line*(a[i+1].x-a[i].x)*2;
        sum+=abs(ST[0].m-t2);
        t2=ST[0].m;
    }
    printf("%d\n",sum);
}
int main(){
    int n,x1,x2,y1,y2,i,j,suby,upy;
    while(~scanf("%d",&n)){
        j=0;
        suby=10000;upy=-10000;
        for(i=0;i<n;i++){
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            a[j].x=x1;a[j].y1=y1;a[j].y2=y2;a[j].d=1;
            j++;
            a[j].x=x2;a[j].y1=y1;a[j].y2=y2;a[j].d=0;
            j++;
            if(suby>y1) suby=y1;
            if(upy<y2) upy=y2;
        }
        sort(a,a+j,cmp);
        build(suby,upy,0);
        cal_c(j);
    }
    return 0;
}