學(xué)習(xí)鏈接：http://blog.csdn.net/lwt36/article/details/48908031

學(xué)習(xí)掃描線主要學(xué)習(xí)的是一種掃描的思想，后期可以求解很多問(wèn)題。

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掃描線求矩形周長(zhǎng)并

hdu 1928

Picture

Time Limit: 6000/2000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4795????Accepted Submission(s): 2339

Problem 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.

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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.

Please process to the end of file.

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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.

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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

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Sample Output 228

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Source IOI 1998

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Recommend linle

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1 //#pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include<cstdio> 3 #include<iostream> 4 #include<cstring> 5 #include<algorithm> 6 #include<cmath> 7 #define clr(x) memset(x,0,sizeof(x)) 8 #define MAXN 50010 9 using namespace std; 10 struct edgx 11 { 12 int l,u,x; 13 int d; 14 }edgex[MAXN]; 15 struct edgy 16 { 17 int l,r,y; 18 int d; 19 }edgey[MAXN]; 20 struct seg 21 { 22 int l,r,cov,len; 23 }segt[MAXN<<2]; 24 int cntx,cnty; 25 int x[MAXN],y[MAXN],vec[MAXN]; 26 bool cmpy(edgy a,edgy b) 27 { 28 if(a.y==b.y) return a.d>b.d; 29 return a.y<b.y; 30 } 31 bool cmpx(edgx a,edgx b) 32 { 33 if(a.x==b.x) return a.d>b.d; 34 return a.x<b.x; 35 } 36 void init(int i,int l,int r) 37 { 38 segt[i]=(seg){l,r,0,0}; 39 if(l==r) 40 return ; 41 int mid=(l+r)>>1; 42 init(i<<1,l,mid); 43 init(i<<1|1,mid+1,r); 44 return ; 45 } 46 void pushup(int i) 47 { 48 if(segt[i].cov) 49 { 50 segt[i].len=vec[segt[i].r+1]-vec[segt[i].l]; 51 } 52 else if(segt[i].l==segt[i].r) 53 { 54 segt[i].len=0; 55 } 56 else 57 { 58 segt[i].len=segt[i<<1].len+segt[i<<1|1].len; 59 } 60 return ; 61 } 62 void update(int i,int l,int r,int value) 63 { 64 if(segt[i].l>=l && segt[i].r<=r) 65 { 66 segt[i].cov+=value; 67 pushup(i); 68 return ; 69 } 70 int mid=(segt[i].l+segt[i].r)>>1; 71 if(mid>=r) 72 { 73 update(i<<1,l,r,value); 74 } 75 else if(mid<l) 76 { 77 update(i<<1|1,l,r,value); 78 } 79 else 80 { 81 update(i<<1,l,r,value); 82 update(i<<1|1,l,r,value); 83 } 84 pushup(i); 85 return ; 86 } 87 int main() 88 { 89 int x1,x2,y1,y2,n,m,T,ans,l,r,k; 90 while(scanf("%d",&n)!=EOF) 91 { 92 cntx=0; 93 cnty=0; 94 for(int i=1;i<=n;i++) 95 { 96 scanf("%d%d%d%d",&x1,&y1,&x2,&y2); 97 edgex[++cntx]=(edgx){y1,y2,x1,1}; 98 x[cntx]=x1; 99 edgex[++cntx]=(edgx){y1,y2,x2,-1}; 100 x[cntx]=x2; 101 edgey[++cnty]=(edgy){x1,x2,y1,1}; 102 y[cnty]=y1; 103 edgey[++cnty]=(edgy){x1,x2,y2,-1}; 104 y[cnty]=y2; 105 } 106 n<<=1; 107 ans=0; 108 memcpy(vec,x,sizeof(x)); 109 sort(vec+1,vec+n+1); 110 m=unique(vec+1,vec+n+1)-vec-1; 111 sort(edgey+1,edgey+n+1,cmpy); 112 init(1,1,m); 113 for(int i=1;i<=n;i++) 114 if(edgey[i].l<edgey[i].r) 115 { 116 k=segt[1].len; 117 l=lower_bound(vec+1,vec+m+1,edgey[i].l)-vec; 118 r=lower_bound(vec+1,vec+m+1,edgey[i].r)-vec; 119 update(1,l,r-1,edgey[i].d); 120 ans+=abs(segt[1].len-k); 121 } 122 memcpy(vec,y,sizeof(y)); 123 sort(vec+1,vec+n+1); 124 m=unique(vec+1,vec+n+1)-vec-1; 125 sort(edgex+1,edgex+n+1,cmpx); 126 init(1,1,m); 127 for(int i=1;i<=n;i++) 128 if(edgex[i].l<edgex[i].u) 129 { 130 k=segt[1].len; 131 l=lower_bound(vec+1,vec+m+1,edgex[i].l)-vec; 132 r=lower_bound(vec+1,vec+m+1,edgex[i].u)-vec; 133 update(1,l,r-1,edgex[i].d); 134 ans+=abs(segt[1].len-k); 135 } 136 printf("%d\n",ans); 137 } 138 return 0; 139 }

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hdu 1255 矩陣面積交

覆蓋的面積

Time Limit: 10000/5000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5718????Accepted Submission(s): 2854

Problem Description 給定平面上若干矩形,求出被這些矩形覆蓋過(guò)至少兩次的區(qū)域的面積.

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Input 輸入數(shù)據(jù)的第一行是一個(gè)正整數(shù)T(1<=T<=100),代表測(cè)試數(shù)據(jù)的數(shù)量.每個(gè)測(cè)試數(shù)據(jù)的第一行是一個(gè)正整數(shù)N(1<=N<=1000),代表矩形的數(shù)量,然后是N行數(shù)據(jù),每一行包含四個(gè)浮點(diǎn)數(shù),代表平面上的一個(gè)矩形的左上角坐標(biāo)和右下角坐標(biāo),矩形的上下邊和X軸平行,左右邊和Y軸平行.坐標(biāo)的范圍從0到100000.

注意:本題的輸入數(shù)據(jù)較多,推薦使用scanf讀入數(shù)據(jù).

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Output 對(duì)于每組測(cè)試數(shù)據(jù),請(qǐng)計(jì)算出被這些矩形覆蓋過(guò)至少兩次的區(qū)域的面積.結(jié)果保留兩位小數(shù).

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Sample Input 2 5 1 1 4 2 1 3 3 7 2 1.5 5 4.5 3.5 1.25 7.5 4 6 3 10 7 3 0 0 1 1 1 0 2 1 2 0 3 1

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Sample Output 7.63 0.00

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Author Ignatius.L & weigang Lee

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1 //#pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include<cstdio> 3 #include<iostream> 4 #include<cstring> 5 #include<algorithm> 6 #include<cmath> 7 #define clr(x) memset(x,0,sizeof(x)) 8 #define MAXN 10010 9 using namespace std; 10 struct edg 11 { 12 double l,r,y; 13 int d; 14 }edge[MAXN]; 15 struct seg 16 { 17 int l,r,cov; 18 double len1,len2; 19 }segt[MAXN<<2]; 20 int cnt; 21 double x[MAXN]; 22 bool cmp(edg a,edg b) 23 { 24 if(a.y==b.y) return a.d>b.d; 25 return a.y<b.y; 26 } 27 double max(double a,double b) 28 { 29 return a>b?a:b; 30 } 31 void init(int i,int l,int r) 32 { 33 segt[i]=(seg){l,r,0,0,0}; 34 if(l==r) 35 return ; 36 int mid=(l+r)>>1; 37 init(i<<1,l,mid); 38 init(i<<1|1,mid+1,r); 39 return ; 40 } 41 void pushup(int i) 42 { 43 if(segt[i].cov>=2) 44 { 45 segt[i].len2=segt[i].len1=x[segt[i].r+1]-x[segt[i].l]; 46 } 47 else if(segt[i].cov==1) 48 { 49 segt[i].len1=x[segt[i].r+1]-x[segt[i].l]; 50 if(segt[i].l==segt[i].r) 51 segt[i].len2=0; 52 else 53 segt[i].len2=max(segt[i<<1].len1,segt[i<<1].len2)+max(segt[i<<1|1].len1,segt[i<<1|1].len2); 54 } 55 else 56 { 57 if(segt[i].l==segt[i].r) 58 { 59 segt[i].len1=segt[i].len2=0; 60 } 61 else 62 { 63 segt[i].len2=segt[i<<1].len2+segt[i<<1|1].len2; 64 segt[i].len1=segt[i<<1].len1+segt[i<<1|1].len1; 65 } 66 } 67 return ; 68 } 69 void update(int i,int l,int r,int value) 70 { 71 if(segt[i].l>=l && segt[i].r<=r) 72 { 73 segt[i].cov+=value; 74 pushup(i); 75 return ; 76 } 77 int mid=(segt[i].l+segt[i].r)>>1; 78 if(mid>=r) 79 { 80 update(i<<1,l,r,value); 81 } 82 else if(mid<l) 83 { 84 update(i<<1|1,l,r,value); 85 } 86 else 87 { 88 update(i<<1,l,r,value); 89 update(i<<1|1,l,r,value); 90 } 91 pushup(i); 92 return ; 93 } 94 int main() 95 { 96 int T,n,m,k,u,v; 97 double x1,x2,y1,y2,ans,l,r; 98 scanf("%d",&T); 99 while(T--) 100 { 101 scanf("%d",&n); 102 cnt=0; 103 ans=0; 104 for(int i=1;i<=n;i++) 105 { 106 scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); 107 edge[++cnt]=(edg){x1,x2,y1,1}; 108 x[cnt]=x1; 109 edge[++cnt]=(edg){x1,x2,y2,-1}; 110 x[cnt]=x2; 111 } 112 n<<=1; 113 sort(x+1,x+n+1); 114 m=unique(x+1,x+n+1)-x-1; 115 sort(edge+1,edge+n+1,cmp); 116 init(1,1,m); 117 for(int i=1;i<n;i++) 118 if(edge[i].r>edge[i].l) 119 { 120 l=lower_bound(x+1,x+m+1,edge[i].l)-x; 121 r=lower_bound(x+1,x+m+1,edge[i].r)-x; 122 update(1,l,r-1,edge[i].d); 123 ans+=segt[1].len2*(edge[i+1].y-edge[i].y); 124 } 125 printf("%0.2lf\n",ans); 126 } 127 return 0; 128 }

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hdu 1542 [POJ 1151]?區(qū)間面積并

Atlantis

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12537????Accepted Submission(s): 5257

Problem Description There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.

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Input The input file consists of several test cases. Each test case starts with a line containing a single integer n (1<=n<=100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0<=x1<x2<=100000;0<=y1<y2<=100000), not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area.

The input file is terminated by a line containing a single 0. Don’t process it.

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Output For each test case, your program should output one section. The first line of each section must be “Test case #k”, where k is the number of the test case (starting with 1). The second one must be “Total explored area: a”, where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.

Output a blank line after each test case.

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Sample Input 2 10 10 20 20 15 15 25 25.5 0

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Sample Output Test case #1 Total explored area: 180.00

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Source Mid-Central European Regional Contest 2000

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卡格式，不說(shuō)了，都是淚。 1 //#pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include<cstdio> 3 #include<iostream> 4 #include<cstring> 5 #include<algorithm> 6 #include<cmath> 7 #define clr(x) memset(x,0,sizeof(x)) 8 #define MAXN 10010 9 using namespace std; 10 struct edg 11 { 12 double l,r,y; 13 int d; 14 }edge[MAXN]; 15 struct seg 16 { 17 int l,r,cov; 18 double len; 19 }segt[MAXN<<2]; 20 int cnt; 21 double x[MAXN]; 22 bool cmp(edg a,edg b) 23 { 24 if(a.y==b.y) return a.d>b.d; 25 return a.y<b.y; 26 } 27 double max(double a,double b) 28 { 29 return a>b?a:b; 30 } 31 void init(int i,int l,int r) 32 { 33 segt[i]=(seg){l,r,0,0}; 34 if(l==r) 35 return ; 36 int mid=(l+r)>>1; 37 init(i<<1,l,mid); 38 init(i<<1|1,mid+1,r); 39 return ; 40 } 41 void pushup(int i) 42 { 43 if(segt[i].cov) 44 { 45 segt[i].len=x[segt[i].r+1]-x[segt[i].l]; 46 } 47 else if(segt[i].l==segt[i].r) 48 { 49 segt[i].len=0; 50 } 51 else 52 { 53 segt[i].len=segt[i<<1].len+segt[i<<1|1].len; 54 } 55 return ; 56 } 57 void update(int i,int l,int r,int value) 58 { 59 if(segt[i].l>=l && segt[i].r<=r) 60 { 61 segt[i].cov+=value; 62 pushup(i); 63 return ; 64 } 65 int mid=(segt[i].l+segt[i].r)>>1; 66 if(mid>=r) 67 { 68 update(i<<1,l,r,value); 69 } 70 else if(mid<l) 71 { 72 update(i<<1|1,l,r,value); 73 } 74 else 75 { 76 update(i<<1,l,r,value); 77 update(i<<1|1,l,r,value); 78 } 79 pushup(i); 80 return ; 81 } 82 int main() 83 { 84 int T,n,m,k,u,v; 85 double x1,x2,y1,y2,ans,l,r; 86 int kase=0; 87 while(scanf("%d",&n) && n!=0) 88 { 89 printf("Test case #%d\n",++kase); 90 cnt=0; 91 ans=0; 92 for(int i=1;i<=n;i++) 93 { 94 scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); 95 edge[++cnt]=(edg){x1,x2,y1,1}; 96 x[cnt]=x1; 97 edge[++cnt]=(edg){x1,x2,y2,-1}; 98 x[cnt]=x2; 99 } 100 n<<=1; 101 sort(x+1,x+n+1); 102 m=unique(x+1,x+n+1)-x-1; 103 sort(edge+1,edge+n+1,cmp); 104 init(1,1,m); 105 for(int i=1;i<n;i++) 106 if(edge[i].r>edge[i].l) 107 { 108 l=lower_bound(x+1,x+m+1,edge[i].l)-x; 109 r=lower_bound(x+1,x+m+1,edge[i].r)-x; 110 update(1,l,r-1,edge[i].d); 111 ans+=segt[1].len*(edge[i+1].y-edge[i].y); 112 } 113 printf("Total explored area: %0.2lf\n",ans); 114 printf("\n"); 115 } 116 return 0; 117 }

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轉(zhuǎn)載于:https://www.cnblogs.com/wujiechao/p/6730638.html

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