扫描线
扫描线求矩形面积并。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <vector> 7 8 using namespace std; 9 10 #define LL long long 11 #define eps 1e-8 12 #define inf 0x3f3f3f3f 13 #define lson l, m, rt << 1 14 #define rson m+1, r, rt << 1 | 1 15 #define mnx 1100 16 17 double x[mnx], sum[mnx]; 18 int vis[mnx]; 19 struct edge{ 20 double l, r, h; 21 int s; 22 edge(){} 23 edge( double l, double r, double h, int s ) : l(l), r(r), h(h), s(s) {} 24 bool operator < ( const edge &b ) const { 25 return h < b.h; 26 } 27 }e[mnx]; 28 void pushup( int rt, int l, int r ){ 29 if( vis[rt] ) sum[rt] = x[r+1] - x[l]; 30 else if( l == r ) sum[rt] = 0; 31 else sum[rt] = sum[rt<<1] + sum[rt<<1|1]; 32 } 33 void update( int L, int R, int v, int l, int r, int rt ){ 34 if( L <= l && R >= r ){ 35 vis[rt] += v; 36 pushup( rt, l, r ); 37 return ; 38 } 39 int m = ( l + r ) >> 1; 40 if( L <= m ) update( L, R, v, lson ); 41 if( R > m ) update( L, R, v, rson ); 42 pushup( rt, l, r ); 43 } 44 int bin( double v, int L, int R ){ 45 int l = L, r = R; 46 while( l < r ){ 47 int m = ( l + r ) >> 1; 48 if( x[m] >= v ) r = m; 49 else l = m + 1; 50 } 51 return l; 52 } 53 int main(){ 54 int n, m, k, kk = 0; 55 while( scanf( "%d", &n ) != EOF && n ){ 56 printf( "Test case #%d\n", ++kk ); 57 memset( vis, 0, sizeof vis ); 58 for( int i = 0; i < mnx; ++i ) 59 sum[i] = 0; 60 m = 0, k = 1; 61 for( int i = 0; i < n; ++i ){ 62 double ax, ay, bx, by; 63 scanf( "%lf %lf %lf %lf", &ax, &ay, &bx, &by ); 64 x[m] = ax, e[m++] = edge( ax, bx, ay, 1 ); 65 x[m] = bx, e[m++] = edge( ax, bx, by, -1 ); 66 } 67 sort( x, x + m ); 68 sort( e, e + m ); 69 for( int i = 1; i < m; ++i ) 70 if( x[i] != x[i-1] ) x[k++] = x[i]; 71 double ans = 0; 72 for( int i = 0; i < m-1; ++i ){ 73 int l = bin( e[i].l, 0, k-1 ); 74 int r = bin( e[i].r, 0, k-1 ) - 1; 75 if( l <= r ) update( l, r, e[i].s, 0, k-1, 1 ); 76 ans += sum[1] * ( e[i+1].h - e[i].h ); 77 } 78 printf( "Total explored area: %.2lf\n\n", ans ); 79 } 80 return 0; 81 }
扫描线求矩形周长并。思路:与面积不同的地方是还要记录竖的边有几个(numseg记录),并且当边界重合的时候需要合并(用lbd和rbd表示边界来辅助)
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <cmath> 6 #include <queue> 7 8 using namespace std; 9 10 #define LL long long 11 #define eps 1e-8 12 #define lson l, m, rt << 1 13 #define rson m + 1, r, rt << 1 | 1 14 #define inf 0x3f3f3f3f 15 #define mnx 23000 16 17 int len[mnx<<2], cnt[mnx<<2], segsum[mnx<<2]; 18 bool lbd[mnx<<2], rbd[mnx<<2]; 19 struct edge{ 20 int l, r, h, s; 21 edge(){} 22 edge( int l, int r, int h, int s ) : l(l), r(r), h(h), s(s) {} 23 bool operator < ( const edge &b ) const { 24 return h < b.h || ( h == b.h && s > b.s ); 25 } 26 }e[mnx]; 27 void pushup( int rt, int l, int r ){ 28 if( cnt[rt] ){ 29 len[rt] = r - l + 1; 30 segsum[rt] = 2; 31 lbd[rt] = rbd[rt] = 1; 32 } 33 else if( l == r ){ 34 len[rt] = segsum[rt] = lbd[rt] = rbd[rt] = 0; 35 } 36 else{ 37 len[rt] = len[rt<<1] + len[rt<<1|1]; 38 segsum[rt] = segsum[rt<<1] + segsum[rt<<1|1]; 39 lbd[rt] = lbd[rt<<1]; 40 rbd[rt] = rbd[rt<<1|1]; 41 if( rbd[rt<<1] && lbd[rt<<1|1] ) segsum[rt] -= 2; 42 } 43 } 44 void update( int L, int R, int v, int l, int r, int rt ){ 45 if( L <= l && R >= r ){ 46 cnt[rt] += v; 47 pushup( rt, l, r ); 48 return ; 49 } 50 int m = ( l + r ) >> 1; 51 if( L <= m ) update( L, R, v, lson ); 52 if( R > m ) update( L, R, v, rson ); 53 pushup( rt, l, r ); 54 } 55 int main(){ 56 int n, m; 57 while( scanf( "%d", &n ) != EOF ){ 58 m = 0; 59 int lmin = inf, rmax = -inf; 60 for( int i = 0; i < n; ++i ){ 61 int ax, ay, bx, by; 62 scanf( "%d%d%d%d", &ax, &ay, &bx, &by ); 63 e[m++] = edge( ax, bx, ay, 1 ); 64 e[m++] = edge( ax, bx, by, -1 ); 65 lmin = min( lmin, ax ); 66 rmax = max( rmax, bx ); 67 } 68 sort( e, e + m ); 69 int ans = 0, pre = 0; 70 for( int i = 0; i < m; ++i ){ 71 if( e[i].l < e[i].r ) 72 update( e[i].l, e[i].r-1, e[i].s, lmin, rmax-1, 1 ); 73 ans += segsum[1] * ( e[i+1].h - e[i].h ); 74 ans += abs( len[1] - pre ); 75 pre = len[1]; 76 } 77 printf( "%d\n", ans ); 78 } 79 return 0; 80 }
POJ 1389 Area of Simple Polygons
矩形面积并。
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <cmath> 6 #include <queue> 7 8 using namespace std; 9 10 #define LL long long 11 #define eps 1e-8 12 #define lson l, m, rt << 1 13 #define rson m + 1, r, rt << 1 | 1 14 #define inf 0x3f3f3f3f 15 #define mnx 53000 16 17 struct edge{ 18 int l, r, h, s; 19 edge () {} 20 edge( int l, int r, int h, int s ) : l(l), r(r), h(h), s(s) {} 21 bool operator < ( const edge &b ) const { 22 return h < b.h || ( h == b.h && s > b.s ); 23 } 24 }e[mnx]; 25 int cnt[mnx<<2], len[mnx<<2]; 26 void pushup( int rt, int l, int r ){ 27 if( cnt[rt] ){ 28 len[rt] = r - l + 1; 29 } 30 else if( l == r ) 31 len[rt] = 0; 32 else 33 len[rt] = len[rt<<1] + len[rt<<1|1]; 34 } 35 void update( int L, int R, int v, int l, int r, int rt ){ 36 if( L <= l && R >= r ){ 37 cnt[rt] += v; 38 pushup( rt, l, r ); 39 return ; 40 } 41 int m = ( l + r ) >> 1; 42 if( L <= m ) update( L, R, v, lson ); 43 if( R > m ) update( L, R, v, rson ); 44 pushup( rt, l, r ); 45 } 46 int main(){ 47 int ax, ay, bx, by; 48 while( scanf( "%d%d%d%d", &ax, &ay, &bx, &by ) != EOF ){ 49 if( ax == -1 && ay == -1 && bx == -1 && by == -1 ) break; 50 int m = 0; 51 e[m++] = edge( ax, bx, ay, 1 ); 52 e[m++] = edge( ax, bx, by, -1 ); 53 while( scanf( "%d%d%d%d", &ax, &ay, &bx, &by ) ){ 54 if( ax == -1 && ay == -1 && bx == -1 && by == -1 ) break; 55 e[m++] = edge( ax, bx, ay, 1 ); 56 e[m++] = edge( ax, bx, by, -1 ); 57 } 58 sort( e, e + m ); 59 int ans = 0; 60 for( int i = 0; i < m; ++i ){ 61 if( e[i].l < e[i].r ) 62 update( e[i].l, e[i].r-1, e[i].s, 0, 50000, 1 ); 63 //printf( "%d %d %d\n", len[1], e[i+1].h, e[i].h ); 64 ans += len[1] * ( e[i+1].h - e[i].h ); 65 } 66 printf( "%d\n", ans ); 67 } 68 return 0; 69 }
fzu 2187 回家种地
求只覆盖了一次的矩形的面积。思路，也是矩形面积并。记录覆盖一次以上sum[]和覆盖两次以上more[]的区间长度，然后sum[1] - more[1]就是覆盖一次的区间长度。
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 #include<cmath> 6 #include<vector> 7 #include<queue> 8 9 using namespace std; 10 11 #define inf 1e16 12 #define eps 1e-6 13 #define LL long long 14 #define ULL unsigned long long 15 #define MP make_pair 16 #define pb push_back 17 #define mod 1000000009 18 #define lson l, m, rt<<1 19 #define rson m+1, r, rt<<1|1 20 #define mnx 200050 21 22 struct edge{ 23 int ax, bx, y, s; 24 edge( int ax = 0, int bx = 0, int y = 0, int s = 0 ) : ax(ax), bx(bx), y(y), s(s) {} 25 bool operator < ( const edge &b ) const { 26 return y < b.y || y == b.y && s > b.s; 27 } 28 }e[mnx]; 29 int x[mnx], vis[mnx<<1]; 30 LL sum[mnx<<1], more[mnx<<1]; 31 void pushup( int rt, int l, int r ){ 32 if( vis[rt] >= 2 ){ 33 sum[rt] = more[rt] = x[r+1] - x[l]; 34 } 35 else if( vis[rt] == 1 ){ 36 sum[rt] = x[r+1] - x[l]; 37 if( l == r ) more[rt] = 0; 38 else more[rt] = sum[rt<<1] + sum[rt<<1|1]; 39 } 40 else{ 41 if( l == r ) sum[rt] = more[rt] = 0; 42 else{ 43 sum[rt] = sum[rt<<1] + sum[rt<<1|1]; 44 more[rt] = more[rt<<1] + more[rt<<1|1]; 45 } 46 } 47 } 48 void update( int L, int R, int s, int l, int r, int rt ){ 49 if( L <= l && R >= r ){ 50 vis[rt] += s; 51 pushup( rt, l, r ); 52 return ; 53 } 54 int m = ( l + r ) >> 1; 55 if( L <= m ) update( L, R, s, lson ); 56 if( R > m ) update( L, R, s, rson ); 57 pushup( rt, l, r ); 58 } 59 int bin( int val, int l, int r ){ 60 while( l < r ){ 61 int m = ( l + r ) >> 1; 62 if( x[m] >= val ) r = m; 63 else l = m+1; 64 } 65 return l; 66 } 67 int main(){ 68 int cas, kk = 1; 69 scanf( "%d", &cas ); 70 while( cas-- ){ 71 memset( vis, 0, sizeof(vis) ); 72 memset( sum, 0, sizeof(sum) ); 73 memset( more, 0, sizeof(more) ); 74 int n; 75 scanf( "%d", &n ); 76 for( int i = 0; i < n; ++i ){ 77 int ax, ay, bx, by; 78 scanf( "%d%d%d%d", &ax, &ay, &bx, &by ); 79 x[i] = ax, e[i] = edge( ax, bx, ay, 1 ); 80 x[i+n] = bx, e[i+n] = edge( ax, bx, by, -1 ); 81 } 82 n <<= 1; 83 sort( x, x + n ); 84 sort( e, e + n ); 85 int m = unique( x, x + n ) - x; 86 LL ans = 0; 87 for( int i = 0; i < n-1; ++i ){ 88 int l = bin( e[i].ax, 0, m-1 ); 89 int r = bin( e[i].bx, 0, m-1 ) - 1; 90 if( l <= r ) update( l, r, e[i].s, 0, m-1, 1 ); 91 //cout << sum[1] << " " << more[1] << endl; 92 ans += (LL)( sum[1] - more[1] ) * ( e[i+1].y - e[i].y ); 93 } 94 printf( "Case %d: %I64d\n", kk++, ans ); 95 } 96 return 0; 97 }
