FJNU 1151 Fat Brother And Geometry(胖哥与几何)

FJNU 1151 Fat Brother And Geometry(胖哥与几何)

Time Limit: 1000MS   Memory Limit: 257792K

【Description】

【题目描述】

Fat brother enrolled in computer graphics, recently teacher arranged a job: given two straight lines and a circle, judge if the round is between straight two lines. If it is, fat brother need to submit a “Yes” to the teacher. If not, fat brother need to submit “No” to the teacher.

Note:

Straight line will be expressed through two point in it

Circle will be expressed through a center point and a radius

胖哥选修了计算机图形学,最近老师给他布置了个任务: 给定两条直线与一个圆,判断这个圆是否在两条直线中。如果是,提交“Yes”给老师。否则,胖哥需要提交“No”给老师。

注意:

直线以两点表示。

圆以圆心和半径表示。

【Input】

【输入】

There are multiple test cases. The first line of input contains an integer T (T <= 20) indicating the number of test cases. For each test case:

The first line contains four number Ax1, Ay1, Ax2, Ay2 denoting there are two point (Ax1, Ay1), (Ax2, Ay2) on the line A

The second line contains four number Bx1, By1, Bx2, By2 denoting there are two point (Bx1, By1), (Bx2, By2) on the line B

The third line contains three number Ox, Oy, r denoting the center of the circle O is (Ox, Oy) and the radius of the circle O is r

( -10^9 <= x, y, r <= 10 ^ 9)

多组测试用例。

第一行是一个整数T(T <= 20)表示测试用例的数量。对于每个测试用例:

第一行有四个数Ax1, Ay1, Ax2, Ay2表示直线A上的两点(Ax1, Ay1), (Ax2, Ay2)。

第二行有四个数Bx1, By1, Bx2, By2表示直线A上的两点(Bx1, By1), (Bx2, By2)。

第行有三个数Ox, Oy, r表示圆心O的坐标(Ox, Oy)和圆O的半径为r。

( -10^9 <= x, y, r <= 10 ^ 9)

【Output】

【输出】

For each test case, output “Yes” or “No”

对于测试用例,输出“Yes”或“No”。

【Sample Input - 输入样例】

【Sample Output - 输出样例】

2

-5.98 5.98 -8.52 2.01

-2.01 4.97 -6.03 -1.06

-6.96 3.94 1.03

-19.98 11.93 19.95 -2.01

16.01 11.93 -17.97 -12.12

12.04 3.92 2.09

No

Yes

【Hint - 提示】

测试用例一

The first sample test case:

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mxeX5fZjeJH46LDY0bGsn9Rp6GLf5YNX5BYin42Nlm/36/VqPlZ/PJM8nsmqwj72CjWaxYNBHl8rBAMmP5AqD0TFrUhXGsdjjzTvZWdhvz5+dVDLfmY9XJsrF2SnM73OICfTqU9bl9bL20z4LmVCcq5mGBpMaK79j90VNGrpoIhycXrhhuMi52pAVj6fH6XD527DDWFBTNwwLJjD1JyEZSKBruWuMivzBs2W0qld2795h4XNkFiUuynhm73rdefPfWO2+xLScta061oWgeFkhmbJrVNz199rT4bv2frienmcMrZ/GK4el0+tDBg8kXX/T8xV/U3Xmnx1Ph07MudlbW9fW5jyKWKVxsOgqYgLp5WCCZsTk2xsudzYnvWB+b5pOrK2aRiuFDTz3V+8ADdb/+dSch9EkdAwM7w+HAnj0G7PVKONjFSVrN3HtlwVvIg0qgdB4WSGYsR9Q3NRA6di1UDKfD1+I8THvX7tbWiXPndP6tl5DWmZnahx5yfvazXq/X2GZchBvrWD03NcxGAVv5sidQYZTOwwLJjOVnOPJvzPU57EbpOQOuayksSPR41s5bkNixZ0/nBV0FtOOL5nJNd9994le/WumOr4TGzmMV+lhgDqrnYYFkxubZu7nQxxpiLH2RFqkYnnrttcuLQLCp3H/7txXtFUiGDfKwQDJjOQZ2PsULEhesGK7/8R/nZmbmbczR7ddcY1QbgAzYIA8LJDP2wgBPnJKlAzyn07mS61qWrBje8M1vDn/ve+733iveGL/6av8dd5S5SyAf9sjDAsmM5QM8oyiuGH6lBYntXV2bn3029uabwfffn/utq68+5HROPPmkYe24HD5cNzb8gythmzwskMxYjlN3igfsFqw6D6nLp3hBYjC4yet1XumZE6+8Empq+sgPf+hZvTp5/rxv8+ax0VFNq+THcI75WTipA2Mrim3ysEAyY/nyPcen5kabc0sL0uX8peLXafEFiVTOvsHBaH9/Mpn0eDyVdVXAF2MW+lh2micHYyuCnfKwQDJj1xNymtyw6QbxHVv8tJEtD1ou8yqGl/I6UVEreAJ2Hm8RcsPF5dMsU7yEGqjGY7M8LJDMWBe7Tyy7AIDD3tNf5JVQl4MZFcNXSJpd8VtYjLnxuo1sUbHRZXGAzfKwQDJj3YSk+OJEDrvsbl1eyy6vPqgZFcNXQp4dI7th39MpsWFuMWablY2yH/bLwwLJjHWwN7Se1+mbmOrKqj1lEz6vj5WhaCzpDxRXDJ+3IFEWkvyKwtm0SMW6prtW81tXyxcF1MWWeVggmbGEL+1NsJtiiRuujp4e9W330ahcorHFFcODwU2VbGi5DLNrGwqFrHzrfNrLGgsXNukDpMCWeVggn7FbWXG27S3bhbH0nd23q49V9C2hcEws9rKpFcPLY5wF4NGjcz0AK2QV50cNDMKueVggn7Fedhs4b9TrWO0Q96dJnk963B5RFXURihckmlQxvAxSbECeXZMV1Se0GlZUkRVSNLS2azVj4zwskM9YN19gkGFxUVQbHJ8e99R52Dke72K/Z03F8OUSZ3PChUjsWevRczobxMraXuWwcR4WyGcsYWUoyACpa6wTxg6/Ptzl7yLbFrvVqmUVw5fLKIv3o6/PdQJ16+vYTXf8GMQag73zsEDKd3YjIfXE9x2fqA+ankmntJTb7WYdlH+BpxfnYdq7mloxfFnE2SUN2XXZxOGE2MAicQg3fTYG2+dhgZTGunh9UF44f+gU7YNI02jTib4TZDOvNnjZ52ZxxXArKyQuSZh1sM2Hm8WKYhqJnVknGwIgEhuB7fOwQEpjCV9O0E3a+tqEsalsKp6L+/1+EqPd6CVPTCQyFlcML5EBNkRPOVLx0bjY0Lm1kzxByG4rG2UbqiEPC2R9f/tZj+TOuoObguI2cOHJsL/Nz0azgYvrn4oXJEodhPK8g51iRyE20PjgO+9jE+AnLG2YLaiSPCyQ1VhKhOZd0vls50BqgMZI2s3GzsSCu4PsrX9h4CdLxfAlibHPoPHz44VZYtbBNuOkjjFUSR4WSGwsHbIeII4XHbSb7Un2EN7NBnYHtM9rLDM7LqkYLumCREGWHQjtYDuGO8SGgDtA4wM7qeO3sl32oHrysEBiYwnvgppI20/baDeby+ey57M9qZ72wXayg90SurhiOM1Cljb0yuTZIZA+MnRmiMYEwldNRG6NkDswRWwAVZWHBXIb62anQ2g3SzNk6EiI8G7Wc5fHu9u713MsdXKuYri8CxIJC/ZkO0lvSDcfbBYbaGSgR8SG4l4r22UPqioPC+Q2lvCOqJYEXwoeeOWAuOfdjh/tePYvXwi/frFiuNOpW9rEKxNjfWwumKs/WM8q4PCL19kIdrOR99esWqotDwukN9bJlkBpj2pjbWNfOPAFkY1rv3YgP8uWDXtc8yuGS0SCr3AaIfQjRlxYR/PwSMOIHtNZ7+q2tnHKU4V5WCC9sZR2QgeuznEnfbtvO7SNpNznf8l01f6opu/q20TNJOnIELaYaYKEJkPj0+Ni2+Cdg+6Um105OGZp22xBFeZhgQrGEp4htxGvy/vwpv/zv/b+u9jmrs+7Iw5Sz5Oz19LmzSNF2JmbERKbjolZbsJP5/iJn52agq4rpjrzsEARY+krMsh62p87Pkny77Ita2aS62I92T9tnWhlU8e0pw1a3UjBECG9TNehmSExW0b4eokuN7+YYQQ1E1dK1eZhgSLGEjagHW+YHgrNVQwntz9HamapEid/fbLv2T7tAU2KJfVdrBYkDcMdxzvEFfmE32l+8LZB8jXePKeVrbMHVZuHBcoYyxYkds99sn5z12dfvUnnt4MlA6mB9Ex65LERx5MOUsu7Yt2S9vHzrutJ/ik2m11Y20R1pcNv9oFSJ1l0V5NqzsMCZYy9+Mm6enXvH+7Q7rqj6UiTuE4geSa5+eDmkcYRt8vNkmcLX3tsJnF+3rWTZG/P1h6sFSslCA/D/XX9eodOrpMmtKtMledhgRrGXlIx/Knb9KxG7iCDg4Nf+sSXxFgxk8tQaQfvHPSP8NmdXr4s2Vf5liW4qw42n5SsSdYfqC9U+m/1tEa/FGUTY9vnX28EyqPK87BAAWOLL9Dx025LVAx3s6veW/taXXe5aArN5XP52Xz9cL3X6Y08EvHMeJhI3Xzo6K5Ms1Lc1TzbRcaV6TjaITp8ws+79t3WF9ADLKV3mvLBUQUgDwsUMLa4YvjFBYmeudljX4tvatcUdVWsUkhkErSzpV5HB6POlJONLV283L7PoPFtnhdDHOXGRkj25mz38e7YE7Hi216N3TXmmfawXQ9Kea5YQZCHC8huLHX1ihXDnWxWlorhSrum2qfCk+HCyc94Ok6/gpuCnT/tdDznYII1c8mFus7ltyN7QdQEn0CqI/kf5Pce29v7RK9YfihgnxTbo84hJytKPIETOYaBPFxAdmMLFRIXrhguztN2EX2zHo1GW1pa6LC2ME8bezk2kBoIuAN1j9T5Bn1zynVzkXz8rlxOrvGV4lWCi5pmtwJia5h8vGTcIEnlUsOvDw/0DhSGrISXgIncGvHmvGzg6uK6AoNAHi5GamNLrRjexS80DRGn5hyJjCS3JOmoksZj+hMaVqm39IuGVdoBbr9vu3+/XzulMXsnCTnE76kh8qyLTyDleNwVePmW9Xww7GH99pE3jsSfiBeLSvjiftqv+jU/G9am5VuApTjIw/OQ19jiColtbVuWqBju5t3aOAvJHrdnIjIRz8Wpt4U7PtLsSvtbUU6VqrvRs9HzVQ+VrXBXLiZblne/7rkNmVyGflHzT799Ov5ovDBSLeBY7Wjb0tbqamWz0+N8kmnQiCMHRSAPz0NeYwt30HG7He3tN5f0Oz7+NUDIZuJv9Pvb/IlcYvT0KO0eqXuFZ4lR7rxfpepSA/Nv5JOjycV3Qp/mW+erW1/nW+vT/l5jayRbeLmmqg5rFQF5+HIkNXZo6BS1SjxedoXEAK943EPIZ+jo1+vd6Y22RGlnSy2l48/C8oZ50CcUOuQFoUpTURs+10CHrGyI28371QB3VV9G60CJIA8viIzG0pequGK4x7N22X9C49fotfLVSL18Ptnvat/a3l7Xnl2dperSsejkW5P0v4tYSjMz/aJ+XvfR66irrhoXE/V/zxUKZ7NQUbhaQZCHF0RGYwsv1Uorhmu8s23kA9Q4nygOEYfuCHqC5AY+P+S8eKYneSYpRqpep3duk5gophn5Lf7fDP+VrXzA7CSgoiAPXwnpjK1IxXAHX9YrVvamL+gX5v/NzD2FduVzA9FE0W+5+OmfjbwOuNuAhoBSQB5eBLmMLV6Q2Ni4oSIvlesK65AKp3lwKtVqkIcXQS5jiyuGm/1SyVouqtpAHl4ciYxNpbI9PXNnVqSuGA4qBvLwkkhkbGFBotQVw0ElQR5eElmM3bv3WCqlQsVwUDGQh0tBCmPp2JWOYMVjqSuGg4qBPFwiUhhbyMNut0PeiuGgkiAPl4j1xtIglEhkCM/DNAtZ3RxgAcjDpWOxsfRjtaPjqHhMe1fax1rbHmA+yMPLwmJjC1nI5VrT2bnV2sYAS0AeXhZWGkuz0NDQXMVwwxYkAqVAHl4ulklSvCAxGNzk9TqtagmwCuThMrDMWDp8LWShFV2gA5QFebgMrDE2mTxTWJBIXypkoSoEebg8LDCW5uHm5sPisd/vamzcYH4bgLUgD5eNBcYWFiReUjEcVBPIw2VjtrGLVQwH1QHy8Eow29jCgkSPZ+0CFcOB3UEeXiGmGltcMRwLEqsT5OEVYp6xxQsSl64YDuwI8vDKMc/Y5ubDNBGRZVUMBzYCedgQTDJ2RRXDgS1AHjYEM8wxoGI4UBzkYaMww9jiD1csSKxCkIcNpOLGFlcMpx+uyMNVCPKwgVTWn+IFiZWqGA7kBnnYWCpibDqd7o1E0q++ej5/bTr9aULW4sO1OkEeNhzjjR34/vd7v/vd3efONfCb2nSTf3mb3BSJPI4FiVUI8rDhGGxsMpk81Nk5de5cIfoESP4rNa9oV72GW01VG8jDlcBgY7u//e2WmZl5r8y+2d82dXY2fuMbxu4LyAzycIUw2NjUa69FL9voptvffNPYHQHJQR6uEAYb61izJjsz47x0Y5Zuv/ZaY3cEZAZ5uHIYbGzDrl3DXV2e994r3jhUU9PY0GDsjoC0IA9XFIONbf3Wt2qffXbvq6+258Xtk8nA1VcPO50Tjz9u7I6AtCAPVxTjz+6MTE6GH3xw1f79rmuvTb/7buD228f+4R80DbmoKkAerjTGG0vljDz+OP1Kp9Mul8vwvw+kBXnYBCq4ShG6VhvIwyaAdfnAGJCHzQHGAgNAHjYNGAsMAHnYNGAsWCnIw2YCY8GKQB42GRgLVgTysMnAWFA+yMPmA2NBmSAPWwKMBWWCPGwJMBaUA/KwVcBYsGyQhy0ExoJlgzxsITAWLA/kYWuBsWAZIA9bDowFywB52HJgLCgV5GEZgLGgJJCHJQHGgpJAHpYEGAuWBnlYHmAsWALkYamAsWAJkIelAsaCxUAelg0YC64I8rCEwFhwRZCHJQTGgoVBHpYTGAsWAHlYWmAsWADkYWmBsWA+yMMyA2PBJSAPSw6MBZeAPCw5MBZcBHlYfmAsmAN5WAlgLJgDeVgJYCxgIA+rAowFyMMqAWMB8rBKwNhqB3lYLWBsVYM8rBwwtqpBHlYOGFu9IA+rCIytUpCHFQXGVinIw4oCY6sR5GF1gbFVB/Kw0sDYqgN5WGlgbHWBPKw6MLaKQB62ATC2ikAetgEwtlpAHrYHMLYqQB62DTC2KkAetg0w1v4gD9sJGGtzkIdtBoy1OcjDNgPG2hnkYfsBY20L8rAtgbG2BXnYlsBYe4I8bFdgrA1BHrYxMNaGIA/bGBhrN5CH7Q2MtRXIw7YHxtoK5GHbA2PtA/JwNVCSsQcPHty1a1elm2LhHm1wgEvmYRsco2x7tOQASzL22LFjJrfM5D3a4ACXzMM2OEbZ9mjJASIV2wHk4eoBxioP5oerChirPJgfripqmpqaSnleiU8zEJP3qOgBvv32R1544Trx+POffyMUaq70HpeF7fdo/gHW9Pf3L/kk2qxSnmYgJu9R0QOkefgzn3mCENbB0jxMR7CV3uOysP0eLTlApGKFQR6uQmCsqmB+uDqBsUqC+eGqpSRj16xZU+l2WLtH5Q6wjDys3DHKv0dLDrAkYx977LFKN8XaPap1gOXlYbWOUYk9WnKASMWKgTxc5cBYxcD8cJUDY1UC88MAxirDyvPw2O5Vf/V/xcO/ef6DA7VGtg6YBIxVhhXn4elfnvpyz78eb8HIV2VgrBoYkIenD/+/f97wHeiqODBWAYyZH/7Xn//zl0/93apVf0UQihUGxipAGXl41apV87Z8QPnU9PS6dVT36d4tu8dq4ayKwFjZKS8PUz0X2Mp1FZz65TSpRURWDxgrNUaulxjbvSruL4ThDZ+GrkoCY6XGyPUStQeej8+FZTZljEisJjBWXgxfL1F74IMPDqy0VcBaYKykYP0wWBAYKylYPwwWBMbKCNYPgysBY6UDeRgsAoyVDuRhsAgwVi6Qh8HiwFiJQB4GSwJjJQJ5GCwJjJUF5GFQCjBWCpCHQYnAWClAHgYlAmOtB3kYlA6MtRjkYbAsYKzFIA+DZQFjrQR5GCwXGGsZyMOgDGCsZSAPgzKAsaaSzWa7H344NTV11dX60ZfWEsL6VeRhUDow1jzGf/KT0N137z57to6qS8h7REuTdf7A3yEPg9KBsSaRyWQ67rln6uxZ/cKWRpK/Z9X0uk++SkidhQ0DagFjTaJ3376dv/mNfunG732Qv/GxSHv4u5Y0CagIjDUJOna9vCfV6ddVV9Hu1+l0mt4ioCQw1iQc11+fTaUu357N5x0Oh+nNAaoCY02i4d57Dx0/3vjOO8Ub44T4brlF0zBRDEoFxpqE3+8f3rIlNDkZee89Ieg4IeFPfGLsmWcsbhlQChhrHoOHD+/t6vrYo486r7km/e67Po9nbGQEkRgsCxhrKu1dXfQrnU67XC6r2wKUBMZaAHQFZQNjAVAJGAuASsBYAFQCxgKgEjAWAJWAsQCoBIwFQCVgLAAqAWMBUAkYC4BKwFgAVALGAqASMBYAlYCxAKgEjAVAJf4brHlPV85dX1oAAAAASUVORK5CYII=" alt="" width="315" height="313" />

测试用例二

The second sample test case:

FJNU 1151 Fat Brother And Geometry(胖哥与几何)

【题解】

大意是判断圆和直线的关系,不过输入数据是double类型。

由于浮点数的精度,最容易发生问题的地方是在两个数比较是否相等的时候,此时应该算上最低的精度,不过题目也没说精度是多少……按之前看过的类似题目就去10e-8好了。

(一开始条件反射在考虑如何用ax+by+c>0直接判断……想了想发觉没什么卵用)

按直线的位置划分情况:

①两条直线相交:圆心到直线的距离是否大于半径。

②两条直线平行:两条平行线的距离是否大于直径。

接着就是两点式化一般式。

FJNU 1151 Fat Brother And Geometry(胖哥与几何)

得:(y1 - y2)x + (x2 - x1)y + x1y2-x2y1 = 0

因此:a = y1 - y2,b = x2 - x1,c = x1y2-x2y1

点到直线的距离:FJNU 1151 Fat Brother And Geometry(胖哥与几何)

平行线间的距离:FJNU 1151 Fat Brother And Geometry(胖哥与几何)

不过两条直线的a,b不定各自相等,转化一下就好了。

之后就没啥好说的了。

【代码 C++】

 #include<cstdio>
#include <cmath>
#define Zero 10e-8
int main(){
int t, n;
double x1, y1, x2, y2, a1, b1, c1, a2, b2, c2, r, L1, L2;
while (~scanf("%d", &t)){
while (t--){
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
a1 = y1 - y2; b1 = x2 - x1; c1 = x1*y2 - x2*y1;
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
a2 = y1 - y2; b2 = x2 - x1; c2 = x1*y2 - x2*y1;
scanf("%lf%lf%lf", &x1, &y1, &r);
L1 = fabs(a1*x1 + b1*y1 + c1) / sqrt(a1*a1 + b1*b1);
L2 = fabs(a2*x1 + b2*y1 + c2) / sqrt(a2*a2 + b2*b2);
if (L1 < r) puts("No");
else if (L2 < r) puts("No");
else if (fabs(a1*b2 - a2*b1) > Zero) puts("Yes");
else if (fabs(fabs(c1 - c2 / (a2 + b2)*(a1 + b1)) / sqrt(a1*a1 + b1*b1) - L1 - L2) > Zero) puts("No");
else puts("Yes");
}
}
return ;
}
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