计算几何。。首先题意很难懂，多亏了纬哥解释，才懂。。
就是有左边有一条折线，右边有一条折线，两条折线的起点和终点的纵坐标相同，还有一些摄像头，一条线段平行x轴的线段从起点到终点，必须得在两条折线中间，并且不能碰到摄像头，问线段最长的长度
其实不难的，思路很快就有了，先o(n)处理左边折线的点到右边的折线的水平长度以及右边的点到左边的点的长度，然后再求摄像头到左右两条折线的水平长度。。不过wa了，我们很快发现了错误在哪，就是摄像头的纵坐标有可能相同，还有y轴坐标大于或者小于折线的终点和起点的摄像头都不能算进去，不过最后还是写挫了，没有ac，今天终于ac了。。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <map> 7 8 using namespace std; 9 10 #define LL long long 11 #define eps 1e-12 12 #define mnx 100020 13 #define mxe 2000020 14 #define inf 1e12 15 #define Pi acos( -1.0 ) 16 17 //精度 18 int dcmp( double x ){ 19 if( fabs( x ) < eps ) return 0; 20 return x < 0 ? -1 : 1; 21 } 22 // 点 23 struct point{ 24 double x, y; 25 point( double x = 0, double y = 0 ) : x(x), y(y) {} 26 point operator + ( const point &b ) const{ 27 return point( x + b.x, y + b.y ); 28 } 29 point operator - ( const point &b ) const{ 30 return point( x - b.x, y - b.y ); 31 } 32 point operator * ( const double &k ) const{ 33 return point( x * k, y * k ); 34 } 35 point operator / ( const double &k ) const{ 36 return point( x / k, y / k ); 37 } 38 bool operator < ( const point &b ) const{ 39 return dcmp( y - b.y ) < 0 || dcmp( y - b.y ) == 0 && dcmp( x - b.x ) < 0; 40 } 41 bool operator == ( const point &b ) const{ 42 return dcmp( x - b.x ) == 0 && dcmp( y - b.y ) == 0; 43 } 44 double len(){ 45 return sqrt( x * x + y * y ); 46 } 47 void input(){ 48 scanf( "%lf%lf", &x, &y ); 49 } 50 }; 51 typedef point Vector; 52 // 点积 53 double dot( Vector a, Vector b ){ 54 return a.x * b.x + a.y * b.y; 55 } 56 // 叉积 57 double cross( Vector a, Vector b ){ 58 return a.x * b.y - a.y * b.x; 59 } 60 int n, m, q; 61 point a[mnx], b[mnx], c[mnx]; 62 double calc( point p0, point p1, point p2 ){ 63 if(dcmp(p1.y - p2.y) == 0) { 64 return min((p0 - p1).len(), (p0 - p2).len()); 65 } 66 if( p1.y > p2.y ) swap( p1, p2 ); 67 double t = (p0.y - p1.y) / ( p2.y - p1.y ); 68 point v = p1 + ( p2 - p1 ) * t; 69 return ( p0 - v ).len(); 70 } 71 void solve(){ 72 double ans = inf; 73 int j = 0; 74 for( int i = 0; i < n; i++ ){ 75 while( j < m-1 ){ 76 if( dcmp(a[i].y-b[j].y) >= 0 && dcmp(a[i].y-b[j+1].y) <= 0 ){ 77 ans = min( ans, calc( a[i], b[j], b[j+1] ) ); 78 break; 79 } 80 else j++; 81 } 82 } 83 j = 0; 84 for( int i = 0; i < m; i++ ){ 85 while( j < n-1 ){ 86 if( dcmp(b[i].y-a[j].y) >= 0 && dcmp(b[i].y-a[j+1].y) <= 0 ){ 87 ans = min( ans, calc( b[i], a[j], a[j+1] ) ); 88 break; 89 } 90 else j++; 91 } 92 } 93 sort( c, c + q ); 94 int cnt = 0; 95 for( int i = 0; i < q; i++ ){ 96 if( dcmp(c[i].y-a[0].y) < 0 || dcmp(c[i].y-a[n-1].y) > 0 ) continue; 97 c[cnt++] = c[i]; 98 } 99 q = cnt; 100 int i = 0; j = 0; 101 for( int k = 0; k < q; k++ ){ 102 double res = -inf; 103 int cur = k; 104 while( cur < q && dcmp(c[cur+1].y - c[cur].y) == 0 ) cur++; 105 int tmp = cur; 106 107 while( i < n-1 ){ 108 if( dcmp(c[k].y-a[i].y) >= 0 && dcmp(c[k].y-a[i+1].y) <= 0 ) 109 break; 110 else i++; 111 } 112 while( j < m-1 ){ 113 if( dcmp(c[k].y-b[j].y) >= 0 && dcmp(c[k].y-b[j+1].y) <= 0 ) 114 break; 115 else j++; 116 }//cout << k << " " << cur << endl; 117 for( int u = k; u < cur; u++ ){ 118 if( cross( c[u] - a[i], a[i+1] - a[i] ) < 0 ) k++; 119 else break; 120 } 121 for( int u = cur; u > k; u-- ){ 122 if( cross( c[u] - b[j], b[j+1] - b[j] ) > 0 ) cur--; 123 else break; 124 } 125 res = max( res, calc( c[k], a[i], a[i+1] ) ); 126 res = max( res, calc( c[cur], b[j], b[j+1] ) ); 127 //printf( "%.5lf ", res ); 128 for( int u = k; u < cur; u++ ){ 129 res = max( res, ( c[u+1] - c[u] ).len() ); 130 } 131 ans = min( res, ans ); 132 k = tmp; 133 } 134 printf( "%.8lf ", ans ); 135 } 136 int main(){ 137 while( scanf( "%d", &n ) != EOF ){ 138 for( int i = 0; i < n; i++ ) 139 a[i].input(); 140 scanf( "%d", &m ); 141 for( int i = 0; i < m; i++ ) 142 b[i].input(); 143 scanf( "%d", &q ); 144 for( int i = 0; i < q; i++ ) 145 c[i].input(); 146 solve(); 147 } 148 return 0; 149 }
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answer: 2.28571429