poj1066Treasure Hunt（线段相交）
很纠结的找到了所有线段的中点，又很纠结的找到了哪些中点可以直接相连，最后bfs一下求出了最短路。。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 2210 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-10; 15 const double inf = ~0u>>2; 16 vector<int>ed[N]; 17 struct Point 18 { 19 double x,y; 20 Point(double x=0,double y=0):x(x),y(y) {} //构造函数 方便代码编写 21 } p[105],q[N]; 22 int g,dis[N]; 23 bool vis[N]; 24 vector<Point>pi[N]; 25 typedef Point pointt; 26 pointt operator + (Point a,Point b) 27 { 28 return Point(a.x+b.x,a.y+b.y); 29 } 30 pointt operator - (Point a,Point b) 31 { 32 return Point(a.x-b.x,a.y-b.y); 33 } 34 pointt operator * (Point a,double b) 35 { 36 return Point(a.x*b,a.y*b); 37 } 38 pointt operator / (Point a,double b) 39 { 40 return Point(a.x/b,a.y/b); 41 } 42 bool operator < (const Point &a,const Point &b) 43 { 44 return a.x<b.x||(a.x==b.x&&a.y<b.y); 45 } 46 int dcmp(double x) 47 { 48 if(fabs(x)<eps) return 0; 49 else return x<0?-1:1; 50 } 51 double cross(Point a,Point b) 52 { 53 return a.x*b.y-a.y*b.x; 54 } 55 Point intersection1(Point a,Point b,Point c,Point d)//直线交点求解 56 { 57 Point pp = a; 58 double t = ((a.x-c.x)*(c.y-d.y)-(a.y-c.y)*(c.x-d.x))/ 59 ((a.x-b.x)*(c.y-d.y)-(a.y-b.y)*(c.x-d.x)); 60 pp.x+=(b.x-a.x)*t; 61 pp.y+=(b.y-a.y)*t; 62 return pp; 63 } 64 bool seginter(pointt a1,pointt a2,pointt b1,pointt b2) 65 { 66 double c1 = cross(a2-a1,b1-a1),c2 = cross(a2-a1,b2-a1), 67 c3 = cross(b2-b1,a1-b1),c4 = cross(b2-b1,a2-b1); 68 return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0; 69 } 70 void init(int i,int n) 71 { 72 p[i+1].x = 0; 73 p[i+1].y = 0; 74 p[i+2].x = 0; 75 p[i+2].y = 100; 76 p[i+3].x = 100; 77 p[i+3].y = 100; 78 p[i+4].x = 100; 79 p[i+4].y = 0; 80 p[i+1+n] = p[i+2]; 81 p[i+2+n] = p[i+3]; 82 p[i+3+n] = p[i+4]; 83 p[i+4+n] = p[i+1]; 84 85 } 86 int judge(int x) 87 { 88 if(fabs(q[x].x)<eps||fabs(q[x].y)<eps||fabs(q[x].x-100.0)<eps||fabs(q[x].y-100.0)<eps) 89 return 1; 90 return 0; 91 } 92 void bfs(int st) 93 { 94 queue<int>Q; 95 Q.push(st); 96 memset(vis,0,sizeof(vis)); 97 // cut<< 98 for(int i = 0 ; i <= st ; i++) 99 dis[i] = INF; 100 dis[st] = 0; 101 while(!Q.empty()) 102 { 103 int u = Q.front(); 104 Q.pop(); 105 //printf("%.2f %.2f %d\n",q[u].x,q[u].y,dis[u]); 106 if(judge(u)) 107 { 108 printf("Number of doors = %d\n",dis[u]); 109 break; 110 } 111 for(int i =0 ; i < ed[u].size() ; i++) 112 { 113 int v = ed[u][i]; 114 //cout<<v<<" "<<dis[u]<<endl; 115 dis[v] = min(dis[v],dis[u]+1); 116 if(!vis[v]) 117 { 118 vis[v] = 1; 119 Q.push(v); 120 } 121 } 122 } 123 } 124 bool cmp(Point a,Point b) 125 { 126 if(fabs(a.x-b.x)<eps) 127 return a.y<b.y; 128 return a.x<b.x; 129 } 130 int main() 131 { 132 int n,i,j,e; 133 scanf("%d",&n); 134 // while(scanf("%d",&n)!=EOF) 135 // { 136 n+=4; 137 g = 0; 138 for(i = 1; i <= n-4 ; i++) 139 { 140 scanf("%lf%lf%lf%lf",&p[i].x,&p[i].y,&p[i+n].x,&p[i+n].y); 141 } 142 init(n-4,n); 143 for(i = 1; i <= n-4; i++) 144 { 145 for(j = 1; j <= n-4 ; j++) 146 { 147 if(i==j) continue; 148 if(seginter(p[i],p[i+n],p[j],p[j+n])) 149 { 150 pi[i].push_back(intersection1(p[i],p[i+n],p[j],p[j+n])); 151 } 152 } 153 } 154 for(i = 1 ; i <= n-4 ; i++) 155 { 156 if(dcmp(p[i].x)==0) 157 pi[n-3].push_back(p[i]); 158 if(dcmp(p[i+n].x)==0) 159 pi[n-3].push_back(p[i+n]); 160 if(dcmp(p[i].y-100)==0) 161 pi[n-2].push_back(p[i]); 162 if(dcmp(p[i+n].y-100)==0) 163 pi[n-2].push_back(p[i+n]); 164 if(dcmp(p[i].x-100)==0) 165 pi[n-1].push_back(p[i]); 166 if(dcmp(p[i+n].x-100)==0) 167 pi[n-1].push_back(p[i+n]); 168 if(dcmp(p[i].y)==0) 169 pi[n].push_back(p[i]); 170 if(dcmp(p[i+n].y)==0) 171 pi[n].push_back(p[i+n]); 172 } 173 for(i = 1 ; i <= n; i++) 174 { 175 pi[i].push_back(p[i]); 176 pi[i].push_back(p[i+n]); 177 sort(pi[i].begin(),pi[i].end(),cmp); 178 for(j = 0 ; j < pi[i].size()-1 ; j++) 179 { 180 Point tp; 181 Point p1 = pi[i][j],p2= pi[i][j+1]; 182 if(dcmp(p1.x-p2.x)==0&&dcmp(p1.y-p2.y)==0) continue; 183 tp.x = (p1.x+p2.x)/2; 184 tp.y = (p1.y+p2.y)/2; 185 // printf("%d tpx = %.2f tpy = %.2f p1x = %.2f p1y = %.2f p2x = %.2f p2y = %.2f\n",i,tp.x,tp.y,p1.x,p1.y,p2.x,p2.y); 186 q[++g] = tp; 187 } 188 pi[i].clear(); 189 } 190 g++; 191 scanf("%lf%lf",&q[g].x,&q[g].y); 192 for(i = 1; i <= g ; i++) 193 ed[i].clear(); 194 for(i = 1; i <= g ; i++) 195 { 196 // printf("%.2f %.2f\n",q[i].x,q[i].y); 197 for(j = 1; j <= g ; j++) 198 { 199 if(i==j) continue; 200 for(e = 1; e <= n; e++) 201 { 202 if(seginter(q[i],q[j],p[e],p[e+n])) 203 { 204 // if(i==g) 205 // printf("%.2f %.2f e = %d\n",q[j].x,q[j].y,e); 206 break; 207 } 208 } 209 if(e>n) 210 { 211 // if(i==g) 212 // printf("%.2f %.2f %.2f %.2f\n",q[i].x,q[i].y,q[j].x,q[j].y); 213 ed[i].push_back(j); 214 } 215 } 216 } 217 bfs(g); 218 // } 219 return 0; 220 }
感觉这题没那么复杂，交对后搜了下题解，，然后 这题是一个水题，确实水。。
因为这个点的最终目的是要走出去的，边界上墙的点相对于边界上其它的点应该是更优的，所以只需枚举这些点到达起点经过多少堵墙即可。
几何抓不住重点是件很纠结的事。
附优美的简短的跑得飞快的代码
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 2210 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-10; 15 const double inf = ~0u>>2; 16 vector<int>ed[N]; 17 struct Point 18 { 19 double x,y; 20 Point(double x=0,double y=0):x(x),y(y) {} //构造函数 方便代码编写 21 } p[105],q[N]; 22 int g,dis[N]; 23 bool vis[N]; 24 vector<Point>pi[N]; 25 typedef Point pointt; 26 pointt operator + (Point a,Point b) 27 { 28 return Point(a.x+b.x,a.y+b.y); 29 } 30 pointt operator - (Point a,Point b) 31 { 32 return Point(a.x-b.x,a.y-b.y); 33 } 34 pointt operator * (Point a,double b) 35 { 36 return Point(a.x*b,a.y*b); 37 } 38 pointt operator / (Point a,double b) 39 { 40 return Point(a.x/b,a.y/b); 41 } 42 bool operator < (const Point &a,const Point &b) 43 { 44 return a.x<b.x||(a.x==b.x&&a.y<b.y); 45 } 46 int dcmp(double x) 47 { 48 if(fabs(x)<eps) return 0; 49 else return x<0?-1:1; 50 } 51 double cross(Point a,Point b) 52 { 53 return a.x*b.y-a.y*b.x; 54 } 55 bool seginter(pointt a1,pointt a2,pointt b1,pointt b2) 56 { 57 double c1 = cross(a2-a1,b1-a1),c2 = cross(a2-a1,b2-a1), 58 c3 = cross(b2-b1,a1-b1),c4 = cross(b2-b1,a2-b1); 59 return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0; 60 } 61 int main() 62 { 63 int n,i,j,e; 64 scanf("%d",&n); 65 g = 0; 66 for(i = 1; i <= n ; i++) 67 { 68 scanf("%lf%lf%lf%lf",&p[i].x,&p[i].y,&p[i+n].x,&p[i+n].y); 69 } 70 Point tp; 71 scanf("%lf%lf",&tp.x,&tp.y); 72 if(n==0) 73 { 74 puts("Number of doors = 1"); 75 return 0; 76 } 77 int ans = INF; 78 for(i = 1 ; i <= n*2; i++) 79 { 80 int ts = 1; 81 for(j = 1 ; j <= n ; j++) 82 { 83 if(seginter(p[i],tp,p[j],p[j+n])) 84 ts++; 85 } 86 ans = min(ans,ts); 87 } 88 printf("Number of doors = %d\n",ans); 89 return 0; 90 }
