HDU 4009 最小树形图
Transfer water
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 4988 Accepted Submission(s): 1798
Problem Description
XiaoA
lives in a village. Last year flood rained the village. So they decide
to move the whole village to the mountain nearby this year. There is no
spring in the mountain, so each household could only dig a well or build
a water line from other household. If the household decide to dig a
well, the money for the well is the height of their house
multiplies X dollar per meter. If the household decide to build a
water line from other household, and if the height of which supply water
is not lower than the one which get water, the money of one water line
is the Manhattan distance of the two households multiplies Y
dollar per meter. Or if the height of which supply water is
lower than the one which get water, a water pump is needed except the
water line. Z dollar should be paid for one water pump. In
addition,therelation of the households must be considered. Some
households may do not allow some other households build a water line
from there house. Now given the 3‐dimensional position (a, b, c)
of every household the c of which means height, can you calculate
the minimal money the whole village need so that every household has
water, or tell the leader if it can’t be done.
Input
Multiple cases.
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.
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.
Output
One
integer in one line for each case, the minimal money the
whole village need so that every household has water. If the plan
does not exist, print “poor XiaoA” in one line.
Sample Input
2 10 20 30
1 3 2
2 4 1
1 2
2 1 2
0 0 0 0
Sample Output
30
1 #include <iostream> 2 #include <cstring> 3 #include <cmath> 4 #define maxn 1002 5 #define maxm 1000002 6 #define inf 0x3fffffff 7 using namespace std; 8 int x,y,z; 9 struct node{ 10 int x,y,z; 11 }ver[maxn]; 12 struct nope{ 13 int u,v,cost; 14 }E[maxm]; 15 int in[maxn],flag[maxn],vis[maxn],pre[maxn]; 16 int dis(node a,node b){ 17 return abs(a.x-b.x)+abs(a.y-b.y)+abs(a.z-b.z); 18 } 19 20 long long int solve(int root,int nv,int ne){ 21 long long int ans=0; 22 int u,v,i,cnt; 23 while(true){ 24 for(i=0;i<nv;i++){ 25 in[i]=inf; 26 } 27 for(i=0;i<ne;i++){ 28 u=E[i].u; 29 v=E[i].v; 30 if(E[i].cost<in[v]&&u!=v){ 31 in[v]=E[i].cost; 32 pre[v]=u; 33 } 34 } 35 memset(flag,-1,sizeof(flag)); 36 memset(vis,-1,sizeof(vis)); 37 cnt=in[root]=0; 38 for(i=0;i<nv;i++){ 39 ans+=in[i]; 40 v=i; 41 while(vis[v]!=i&&v!=root&&flag[v]==-1){ 42 vis[v]=i; 43 v=pre[v]; 44 } 45 if(v!=root&&flag[v]==-1){ 46 for(u=pre[v];u!=v;u=pre[u]){ 47 flag[u]=cnt; 48 } 49 flag[v]=cnt++; 50 } 51 } 52 if(cnt==0){ 53 return ans; 54 } 55 for(i=0;i<nv;i++){ 56 if(flag[i]==-1){ 57 flag[i]=cnt++; 58 } 59 } 60 for(i=0;i<ne;i++){ 61 v=E[i].v; 62 E[i].u=flag[E[i].u]; 63 E[i].v=flag[v]; 64 if(E[i].u!=E[i].v){ 65 E[i].cost-=in[v]; 66 } 67 } 68 nv=cnt; 69 root=flag[root]; 70 } 71 return ans; 72 } 73 74 int main(int argc, char const *argv[]){ 75 int n,i,a,b,id; 76 cin.sync_with_stdio(false); 77 while(cin>>n>>x>>y>>z&&(n||x||y||z)){ 78 for(i=0;i<n;i++){ 79 cin>>ver[i].x>>ver[i].y>>ver[i].z; 80 } 81 for(i=0,id=0;i<n;i++){ 82 cin>>a; 83 while(a--){ 84 cin>>b; 85 b--; 86 E[id].cost=dis(ver[i],ver[b])*y; 87 if(ver[b].z>ver[i].z){ 88 E[id].cost+=z; 89 } 90 E[id].u=i; 91 E[id++].v=b; 92 } 93 } 94 for(i=0;i<n;i++){ 95 E[id].u=n; 96 E[id].v=i; 97 E[id++].cost=ver[i].z*x; 98 } 99 cout<<solve(n,n+1,id)<<endl; 100 } 101 return 0; 102 }