首页 > 代码库 > POJ1984 Navigation Nightmare
POJ1984 Navigation Nightmare
Time Limit: 2000MS | Memory Limit: 30000K | |
Total Submissions: 6219 | Accepted: 2230 | |
Case Time Limit: 1000MS |
Description
Farmer John‘s pastoral neighborhood has N farms (2 <= N <= 40,000), usually numbered/labeled 1..N. A series of M (1 <= M < 40,000) vertical and horizontal roads each of varying lengths (1 <= length <= 1000) connect the farms. A map of these farms might look something like the illustration below in which farms are labeled F1..F7 for clarity and lengths between connected farms are shown as (n):
Being an ASCII diagram, it is not precisely to scale, of course.
Each farm can connect directly to at most four other farms via roads that lead exactly north, south, east, and/or west. Moreover, farms are only located at the endpoints of roads, and some farm can be found at every endpoint of every road. No two roads cross, and precisely one path
(sequence of roads) links every pair of farms.
FJ lost his paper copy of the farm map and he wants to reconstruct it from backup information on his computer. This data contains lines like the following, one for every road:
There is a road of length 10 running north from Farm #23 to Farm #17
There is a road of length 7 running east from Farm #1 to Farm #17
...
As FJ is retrieving this data, he is occasionally interrupted by questions such as the following that he receives from his navigationally-challenged neighbor, farmer Bob:
What is the Manhattan distance between farms #1 and #23?
FJ answers Bob, when he can (sometimes he doesn‘t yet have enough data yet). In the example above, the answer would be 17, since Bob wants to know the "Manhattan" distance between the pair of farms.
The Manhattan distance between two points (x1,y1) and (x2,y2) is just |x1-x2| + |y1-y2| (which is the distance a taxicab in a large city must travel over city streets in a perfect grid to connect two x,y points).
When Bob asks about a particular pair of farms, FJ might not yet have enough information to deduce the distance between them; in this case, FJ apologizes profusely and replies with "-1".
F1 --- (13) ---- F6 --- (9) ----- F3
| |
(3) |
| (7)
F4 --- (20) -------- F2 |
| |
(2) F5
|
F7
Being an ASCII diagram, it is not precisely to scale, of course.
Each farm can connect directly to at most four other farms via roads that lead exactly north, south, east, and/or west. Moreover, farms are only located at the endpoints of roads, and some farm can be found at every endpoint of every road. No two roads cross, and precisely one path
(sequence of roads) links every pair of farms.
FJ lost his paper copy of the farm map and he wants to reconstruct it from backup information on his computer. This data contains lines like the following, one for every road:
There is a road of length 10 running north from Farm #23 to Farm #17
There is a road of length 7 running east from Farm #1 to Farm #17
...
As FJ is retrieving this data, he is occasionally interrupted by questions such as the following that he receives from his navigationally-challenged neighbor, farmer Bob:
What is the Manhattan distance between farms #1 and #23?
FJ answers Bob, when he can (sometimes he doesn‘t yet have enough data yet). In the example above, the answer would be 17, since Bob wants to know the "Manhattan" distance between the pair of farms.
The Manhattan distance between two points (x1,y1) and (x2,y2) is just |x1-x2| + |y1-y2| (which is the distance a taxicab in a large city must travel over city streets in a perfect grid to connect two x,y points).
When Bob asks about a particular pair of farms, FJ might not yet have enough information to deduce the distance between them; in this case, FJ apologizes profusely and replies with "-1".
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains four space-separated entities, F1,
F2, L, and D that describe a road. F1 and F2 are numbers of
two farms connected by a road, L is its length, and D is a
character that is either ‘N‘, ‘E‘, ‘S‘, or ‘W‘ giving the
direction of the road from F1 to F2.
* Line M+2: A single integer, K (1 <= K <= 10,000), the number of FB‘s
queries
* Lines M+3..M+K+2: Each line corresponds to a query from Farmer Bob
and contains three space-separated integers: F1, F2, and I. F1
and F2 are numbers of the two farms in the query and I is the
index (1 <= I <= M) in the data after which Bob asks the
query. Data index 1 is on line 2 of the input data, and so on.
Output
* Lines 1..K: One integer per line, the response to each of Bob‘s
queries. Each line should contain either a distance
measurement or -1, if it is impossible to determine the
appropriate distance.
Sample Input
7 61 6 13 E6 3 9 E3 5 7 S4 1 3 N2 4 20 W4 7 2 S31 6 11 4 32 6 6
Sample Output
13-110
Hint
At time 1, FJ knows the distance between 1 and 6 is 13.
At time 3, the distance between 1 and 4 is still unknown.
At the end, location 6 is 3 units west and 7 north of 2, so the distance is 10.
At time 3, the distance between 1 and 4 is still unknown.
At the end, location 6 is 3 units west and 7 north of 2, so the distance is 10.
Source
USACO 2004 February
偏移量并查集。
用偏移量保存点和点之间的相对距离,以祖先节点为基准就可以知道两点之间的坐标差。
刚开始想着不压缩路径会更好写,其实不然……
WA到飞起,最后把求相对距离的算式乘了-1,迷一样地过了
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 using namespace std; 7 const int mxn=45000; 8 struct quest{ 9 int a,b;10 int id;11 int ti;12 }que[mxn];13 int cmp(const quest a,const quest b){14 return a.ti<b.ti;15 }16 int ans[mxn];17 struct node{18 int x,y;19 int fa;20 }p[mxn];21 int find(int x){22 if(p[x].fa==x)return x;23 int tmp=p[x].fa;24 p[x].fa=find(p[x].fa);25 p[x].x+=p[tmp].x;26 p[x].y+=p[tmp].y;27 return p[x].fa;28 }29 int n,m,R;30 int f1[mxn],f2[mxn],di[mxn];//修改信息 31 char dr[mxn];32 /*33 void Print(){34 for(int i=1;i<=n;i++){35 printf("i:%d pos: %d,%d fa:%d\n",i,p[i].x,p[i].y,p[i].fa);36 }37 return;38 }*/39 int main(){40 scanf("%d%d",&n,&m);41 int i,j;42 for(i=1;i<=n;i++) p[i].fa=i;43 for(i=1;i<=m;i++){44 scanf("%d%d%d %c",&f1[i],&f2[i],&di[i],&dr[i]);45 }46 scanf("%d",&R);47 for(i=1;i<=R;i++){scanf("%d%d%d",&que[i].a,&que[i].b,&que[i].ti);que[i].id=i;}48 sort(que+1,que+R+1,cmp);49 int tmm=1;50 for(i=1;i<=R;i++){51 while(tmm<=m && tmm<=que[i].ti){52 int fa1=find(f1[tmm]);53 int fa2=find(f2[tmm]);54 switch(dr[tmm]){55 case ‘E‘:{56 p[fa2].fa=fa1;57 p[fa2].x=-p[f2[tmm]].x+p[f1[tmm]].x-di[tmm];58 p[fa2].y=-p[f2[tmm]].y+p[f1[tmm]].y;59 break;60 }61 case ‘W‘:{62 p[fa2].fa=fa1;63 p[fa2].x=-p[f2[tmm]].x+p[f1[tmm]].x+di[tmm];64 p[fa2].y=-p[f2[tmm]].y+p[f1[tmm]].y;65 break;66 }67 case ‘N‘:{68 p[fa2].fa=fa1;69 p[fa2].x=-p[f2[tmm]].x+p[f1[tmm]].x;70 p[fa2].y=-p[f2[tmm]].y+p[f1[tmm]].y-di[tmm];71 break;72 }73 case ‘S‘:{ 74 p[fa2].fa=fa1;75 p[fa2].x=-p[f2[tmm]].x+p[f1[tmm]].x;76 p[fa2].y=-p[f2[tmm]].y+p[f1[tmm]].y+di[tmm];77 break;78 }79 }80 tmm++;81 }82 int fa1=find(que[i].a); int fa2=find(que[i].b);83 if(fa1!=fa2){ans[que[i].id]=-1;}84 else ans[que[i].id]=abs(p[que[i].a].x-p[que[i].b].x)+abs(p[que[i].a].y-p[que[i].b].y);85 }86 for(i=1;i<=R;i++)printf("%d\n",ans[i]);87 return 0;88 }
POJ1984 Navigation Nightmare
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。