标签：lang 答案 lib rest double 解决 scan 预处理 any
Description
Input
Output
Sample Input
0 0 10000 1000
0 200 5000 200 7000 200 -1 -1
2000 600 5000 600 10000 600 -1 -1
Sample Output
21
Source
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #include<cmath> #define ll long long using namespace std; int read() { int x=0,y=1; char ch=getchar(); while(ch<‘0‘||ch>‘9‘) { if(ch==‘-‘) y=-1; ch=getchar(); } while(ch>=‘0‘&&ch<=‘9‘) { x=x*10+ch-‘0‘; ch=getchar(); } return x*y; } int abs(int x) { if(x<0) return -x; else return x; } int change(double x) { int r=(int)x; if(x-r<0.5) return r; else return r+1; } double way(int x1,int y1,int x2,int y2) { int r1=abs(x1-x2),r2=abs(y1-y2); return sqrt(r1*r1+r2*r2); } struct edge { int next,to; double lon; } e[4045]; int map[305][305],num,node[305][3],head[6045],cnt,t[345],headd,tail=1; double dist[345]; bool vis[345]; void add(int from,int to,double lon) { e[++cnt].lon=lon; e[cnt].to=to; e[cnt].next=head[from]; head[from]=cnt; } int main() { memset(head,-1,sizeof(head)); int x1=read(),y1=read(),x2=read(),y2=read(),x,y,p=1,l=change(way(x1,y1,x2,y2)/1000*6); node[++num][0]=x1; node[num][1]=y1; node[++num][0]=x2; node[num][1]=y2; add(1,2,l); add(2,1,l); while(scanf("%d%d",&x,&y)!=EOF) { if(x==-1&&y==-1) { p++; continue; } node[++num][0]=x; node[num][1]=y; node[num][2]=p; for(int i=1; i<num; i++) { double dis; if(node[i][2]==p) dis=way(x,node[i][0],y,node[i][1])/4000*6; else dis=way(x,node[i][0],y,node[i][1])/1000*6; // printf("x=%d y=%d node[i][0]=%d node[i][1]=%d dis=%f\n",x,y,node[i][0],node[i][1],dis); add(num,i,dis),add(i,num,dis); } } for(int i=2; i<=num; i++) dist[i]=2e8; t[0]=1; while(headd!=tail) { int r=head[t[headd]]; vis[t[headd]]=0; while(r!=-1) { if(dist[e[r].to]>dist[t[headd]]+e[r].lon) { dist[e[r].to]=dist[t[headd]]+e[r].lon; printf("e[r].lon=%f e[r].to=%d dist=%f %f\n",e[r].lon,e[r].to,dist[t[headd]],dist[e[r].to]); if(!vis[e[r].to]) { vis[e[r].to]=1; t[tail++]=e[r].to; } } r=e[r].next; } headd++; } printf("%f",dist[2]); return 0; } // FOR C.H
附上最新经谭姐启发写的正解方法代码（思路稍后，仍有问题在调试）：
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #include<cmath> using namespace std; int head[4045],cnt,num; struct edge { int next,to; double lon; } e[4045]; void add(int from,int to,double lon) { e[++cnt].lon=lon; e[cnt].to=to; e[cnt].next=head[from]; head[from]=cnt; } int abs(int x) { if(x<0) return -x; else return x; } double far(int x1,int y1,int x2,int y2) { int r1=abs(x1-x2),r2=abs(y1-y2); return sqrt(r1*r1+r2*r2); } int change(double x) { int r=(int)x; if(x-r<0.5) return r; else return r+1; } int sub[345][2],map[345][2]; void scan() { int x,y,p=0,s=0; while(scanf("%d%d",&x,&y)!=EOF) { if(x==-1&&y==-1) { for(int i=2; i<=s; i++) { double f=far(map[num-s+i-1][0],map[num-s+i-1][1],map[num-s+i][0],map[num-s+i][1])/4000*6; add(num-s+i-1,num-s+i,f); add(num-s+i,num-s+i-1,f); } s=0; continue; } map[++num][0]=x; map[num][1]=y; s++; } } double dist[345]; int t[345],headd,tail=1; bool vis[345]; int main() { memset(head,-1,sizeof(head)); int x1,y1,x2,y2; scanf("%d%d%d%d",&x1,&y1,&x2,&y2); scan(); for(int i=1; i<=num; i++) for(int j=1; j<=num; j++) { double f=far(map[i][0],map[i][1],map[j][0],map[j][1])/1000*6; add(i,j,f); } num++; for(int i=1; i<num; i++) { double f=far(map[i][0],map[i][1],x1,y1)/1000*6; add(i,num,f); add(num,i,f); } num++; for(int i=1; i<num-1; i++) { double f=far(map[i][0],map[i][1],x2,y2)/1000*6; add(i,num,f); add(num,i,f); } t[0]=num-1; vis[num-1]=1; for(int i=1; i<=num; i++) dist[i]=2e8; dist[num-1]=0; while(headd!=tail) { int r=head[t[headd]]; printf("r=%d\n",r); while(r!=-1) { if(dist[e[r].to]>dist[t[headd]]+e[r].lon) { dist[e[r].to]=dist[t[headd]]+e[r].lon; printf("dist[e[r].to]=%f\n",dist[e[r].to]); if(!vis[e[r].to]) { vis[e[r].to]=1; t[tail++]=e[r].to; } } r=e[r].next; } headd++; } printf("%d",change(dist[num])); return 0; } // FOR C.H
标签：lang 答案 lib rest double 解决 scan 预处理 any
原文地址：http://www.cnblogs.com/gshdyjz/p/7152580.html
