hdu 4009 Transfer water(最小型树图)
Transfer water
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 3995 Accepted Submission(s): 1438
First line of each case contains 4 integers n (1<=n<=1000), the number of the households, X (1<=X<=1000), Y (1<=Y<=1000), Z (1<=Z<=1000).
Each of the next n lines contains 3 integers a, b, c means the position of the i‐th households, none of them will exceeded 1000.
Then next n lines describe the relation between the households. The n+i+1‐th line describes the relation of the i‐th household. The line will begin with an integer k, and the next k integers are the household numbers that can build a water line from the i‐th household.
If n=X=Y=Z=0, the input ends, and no output for that.
1 3 2
2 4 1
1 2
2 1 2
0 0 0 0
In 3‐dimensional space Manhattan distance of point A (x1, y1, z1) and B(x2, y2, z2) is |x2‐x1|+|y2‐y1|+|z2‐z1|.
------------ 遇到一个国人发明的算法(algorithm)--------
下面就开始简单的剖析一下,下面的部分吧! 看图
----------~~~~~~~~~~~~~~~魔板AC~~~~~~~~~~~~~--------------
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define maxn 1010
#define type int
const int inf = ~0u >> ;
struct node
{
int u,v;
type cost;
node(){}
node(int _u,int _v,type _c):u(_u),v(_v),cost(_c){}
}e[maxn * maxn];
int pre[maxn],id[maxn],vis[maxn];
type in[maxn];
type dirmst(int root,int nv,int ne)
{
type ret = ;
while()
{
//find the smallest in-arc
fill(in,in + nv,inf);
for(int i = ;i < ne;i++)
{
int u = e[i].u;
int v = e[i].v;
if(e[i].cost < in[v] && u != v)
{
pre[v] = u;
in[v] = e[i].cost;
}
}
for(int i = ;i < nv;i++)
{
if(i == root)
continue;
if(in[i] == inf)
return -;//there are some nodes other than root with no in-arc connected to it
}
//find the dir circle
int cntnode = ;
fill(id,id + nv,-);
fill(vis,vis + nv,-);
in[root] = ;
for(int i = ;i < nv;i++)
{
ret += in[i];
int v = i;
while(vis[v] != i && id[v] == - && v != root)
{
vis[v] = i;
v = pre[v];
}
if(v != root && id[v] == -)
{
for(int u = pre[v]; u != v;u = pre[u])
id[u] = cntnode;
id[v] = cntnode++;
}
}
if(cntnode == )
break;//no circle
for(int i = ;i < nv;i++)
if(id[i] == -)
id[i] = cntnode++;
//compress the nodes
for(int i = ;i < ne;i++)
{
int v = e[i].v;
e[i].u = id[e[i].u];
e[i].v = id[e[i].v];
if(e[i].u != e[i].v)
e[i].cost -= in[v];
}
nv = cntnode;
root = id[root];
}
return ret;
}
int n,tot,X,Y,Z;
int ab(int x)
{
return x >= ?x:-x;
}
struct point
{
int x,y,z;
point(){}
point(int a,int b,int c):x(a),y(b),z(c){}
point operator - (const point p)
{
return point(x - p.x,y - p.y,z - p.z);
}
int dis()
{
return ab(x) + ab(y) + ab(z);
}
}p[maxn];
int main()
{
while(scanf("%d %d %d %d",&n,&X,&Y,&Z) == && (n || X || Y || Z))
{
tot = ;
for(int i = ;i <= n;i++)
{
int a,b,c;
scanf("%d %d %d",&a,&b,&c);
p[i] = point(a,b,c);
e[tot++] = node(,i,ab(p[i].z) * X);
}
for(int i = ;i <= n;i++)
{
int opt;
scanf("%d",&opt);
for(int j = ;j < opt;j++)
{
int a;
scanf("%d",&a);
if(a == i)
continue;
int temp = Y * (p[i] - p[a]).dis();
if(p[i].z < p[a].z)
temp += Z;
e[tot++] = node(i,a,temp);
}
}
int ans = dirmst(,n + ,tot);
if(ans == -)
puts("poor XiaoA");
else
printf("%d\n",ans);
}
}
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