POJ 1984 Navigation Nightmare

时间:2021-02-16 17:38:52

并查集,给n个点和m条边,每条边有方向和长度,再给q个询问,第i个询问查询两个点之间在Ti时刻时的曼哈顿距离(能连通则输出曼哈顿距离,否则输出-1)

这题跟Corporative Network 有点像,只不过那题是维护到根节点的距离,这题还要顺便维护与根节点的x,y方向的偏移量。findset时,每次找完father就要加上father的x、y坐标偏移量,这样findset完以后就得到了与根的偏移量。然后合并时, (注意,这里是 fa[x] = y)

dr[x].x = r.x - dr[r.u].x + dr[r.v].x;

dr[x].y = r.y - dr[r.u].y + dr[r.v].y;      即 x->y <==> u->v - u->x + v->y 如下图:

POJ 1984 Navigation Nightmare

这题还要注意数据可能不是按照时间顺序输入的,要做一个排序,然后再按原来的顺序输出。

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define N 40100 int fa[N],q; struct DR
{
int x,y;
}dr[N],to[]; struct ROAD
{
int u,v;
int x,y;
}road[N]; struct QUERY
{
int a,b,time;
int id,res;
}query[N]; void makeset(int n)
{
for(int i=;i<=n;i++)
{
fa[i] = i;
dr[i].x = dr[i].y = ;
}
} int findset(int x)
{
if(x != fa[x])
{
int tmp = fa[x];
fa[x] = findset(fa[x]);
dr[x].x += dr[tmp].x;
dr[x].y += dr[tmp].y;
}
return fa[x];
} void unionset(int index) //合并第index条边
{
ROAD r = road[index];
int x = findset(r.u);
int y = findset(r.v);
if(x == y)
return;
fa[x] = y;
dr[x].x = r.x - dr[r.u].x + dr[r.v].x; // x->y <==> u->v - u->x + v->y
dr[x].y = r.y - dr[r.u].y + dr[r.v].y;
} int cmp1(QUERY ka,QUERY kb)
{
return ka.time<kb.time;
} int cmp2(QUERY ka,QUERY kb)
{
return ka.id<kb.id;
} void InitDirection()
{
to['E'-'A'].x = ;
to['E'-'A'].y = ;
to['W'-'A'].x = -;
to['W'-'A'].y = ;
to['N'-'A'].x = ;
to['N'-'A'].y = ;
to['S'-'A'].x = ;
to['S'-'A'].y = -;
} void read()
{
int n,m,i,dis;
char ss[];
InitDirection();
scanf("%d%d",&n,&m);
makeset(n);
for(i=;i<=m;i++)
{
scanf("%d%d%d %s",&road[i].u,&road[i].v,&dis,ss);
road[i].x = dis*(to[ss[]-'A'].x);
road[i].y = dis*(to[ss[]-'A'].y);
}
scanf("%d",&q);
for(i=;i<=q;i++)
{
scanf("%d%d%d",&query[i].a,&query[i].b,&query[i].time);
query[i].id = i;
}
sort(query+,query+q+,cmp1);
} void solve()
{
int i,j;
j=;
for(i=;i<=q;i++)
{
for(;j<=query[i].time;j++)
{
unionset(j);
}
if(findset(query[i].a) != findset(query[i].b))
query[i].res = -;
else
query[i].res = abs(dr[query[i].a].x-dr[query[i].b].x) + abs(dr[query[i].a].y-dr[query[i].b].y);
}
sort(query+,query+q+,cmp2);
for(i=;i<=q;i++)
cout<<query[i].res<<endl;
} int main()
{
read();
solve();
return ;
}