The plan of city rebuild
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 616 Accepted Submission(s): 215
Problem Description
News comes!~City W will be rebuilt with the expectation to become a center city. There are some villages and roads in the city now, however. In order to make the city better, some new villages should be built and some old ones should be destroyed. Then the
officers have to make a new plan, now you , as the designer, have the task to judge if the plan is practical, which means there are roads(direct or indirect) between every two villages(of course the village has not be destroyed), if the plan is available,
please output the minimum cost, or output"what a pity!".
Input
Input contains an integer T in the first line, which means there are T cases, and then T lines follow.
Each case contains three parts. The first part contains two integers l(0<l<100), e1, representing the original number of villages and roads between villages(the range of village is from 0 to l-1), then follows e1 lines, each line contains three integers a, b, c (0<=a, b<l, 0<=c<=1000), a, b indicating the village numbers and c indicating the road cost of village a and village b . The second part first contains an integer n(0<n<100), e2, representing the number of new villages and roads(the range of village is from l to l+n-1), then follows e2 lines, each line contains three integers x, y, z (0<=x, y<l+n, 0<=z<=1000), x, y indicating the village numbers and z indicating the road cost of village x and village y. The third part contains an integer m(0<m<l+n), representing the number of deserted villages, next line comes m integers, p1,p2,…,pm,(0<=p1,p2,…,pm<l+n) indicating the village number.
Pay attention: if one village is deserted, the roads connected are deserted, too.
Each case contains three parts. The first part contains two integers l(0<l<100), e1, representing the original number of villages and roads between villages(the range of village is from 0 to l-1), then follows e1 lines, each line contains three integers a, b, c (0<=a, b<l, 0<=c<=1000), a, b indicating the village numbers and c indicating the road cost of village a and village b . The second part first contains an integer n(0<n<100), e2, representing the number of new villages and roads(the range of village is from l to l+n-1), then follows e2 lines, each line contains three integers x, y, z (0<=x, y<l+n, 0<=z<=1000), x, y indicating the village numbers and z indicating the road cost of village x and village y. The third part contains an integer m(0<m<l+n), representing the number of deserted villages, next line comes m integers, p1,p2,…,pm,(0<=p1,p2,…,pm<l+n) indicating the village number.
Pay attention: if one village is deserted, the roads connected are deserted, too.
Output
For each test case, If all villages can connect with each other(direct or indirect), output the minimum cost, or output "what a pity!".
Sample Input
2 4 5 0 1 10 0 2 20 2 3 40 1 3 10 1 2 70 1 1 4 1 60 2 2 3 3 3 0 1 20 2 1 40 2 0 70 2 3 0 3 10 1 4 90 2 4 100 0
Sample Output
70 160
Author
wangjing
题目大意：先给你n1个点。m1条边。又给你n2个点。m2条边。然后告诉你有些点不须要。然后求剩下点的最
小生成树。
反正光看题目就花了半小时。真不明确了，老路，新路，给边还分两次给。。
用了prim和kruskal两种算法来实现。中间还debug了蛮久。
Prim算法：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int INF =1e8;
int mp[205][205];
int des[205];
int visi[205];
int low[205]; //更新最小值
int n;
int prim()
{
int i,j;
memset(visi,0,sizeof(visi));
for(i=0;i<=200;i++)
low[i]=INF;
int ans=0,pos=-1;
for(i=0;i<n;i++)
{
if(!des[i])
{
pos=i;
break;
}
}
if(pos==-1) return 0;
for(i=0;i<=200;i++)
low[i]=mp[pos][i];
visi[pos]=1;
int mi;
for(i=0;i<n;i++)
{
mi=INF;
int flag=0;
for(j=0;j<n;j++)
{
if(des[j]) continue;
if(!visi[j])
{
flag=1;
if(low[j]<mi)
{
mi=low[j];
pos=j;
}
}
}
if(mi==INF&&flag)
return -1;
if(!flag) return ans;
ans+=mi;
visi[pos]=1;
for(j=0;j<n;j++)
if(!visi[j])
low[j]=min(low[j],mp[pos][j]);
}
return ans;
}
int main()
{
int tes,i,j,res;
scanf("%d",&tes);
while(tes--)
{
int n1,m,u,v,val;
scanf("%d%d",&n1,&m);
for(i=0;i<=200;i++)
for(j=0;j<=200;j++)
mp[i][j]=INF;
memset(des,0,sizeof(des));
for(i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&val);
if(mp[u][v]>val) //这个地方。边可能会多次给出，坑。。
{
mp[u][v]=val;
mp[v][u]=val;
}
}
int n2;
scanf("%d%d",&n2,&m);
for(i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&val);
if(mp[u][v]>val)
{
mp[u][v]=val;
mp[v][u]=val;
}
}
n=n1+n2;
int t,x;
scanf("%d",&t);
for(i=0;i<t;i++)
{
scanf("%d",&x);
des[x]=1;
}
res=prim();
if(res==-1) puts("what a pity!");
else printf("%d
",res);
}
return 0;
}
Kruskal算法：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF =1e6;
const int maxn=205;
int n,m,t; //总点数，总边数，desert的边数
int des[maxn]; //为1代表城市荒废
int fa[maxn];
struct Edge
{
int from;
int to;
int val;
}edge[maxn*maxn];
int cmp(Edge p1,Edge p2)
{
//if(des[p1.from]||des[p1.to]) return 0;
//if(des[p2.from]||des[p2.to]) return 1;
return p1.val<p2.val;
}
void init()
{
for(int i=0;i<=200;i++)
fa[i]=i;
}
int find1(int x)
{
if(fa[x]!=x) fa[x]=find1(fa[x]);
return fa[x];
}
void merge1(int p1,int p2)
{
p1=find1(p1);
p2=find1(p2);
fa[p1]=p2;
}
int Kruskal()
{
sort(edge,edge+m,cmp);
int ans=0,ste=0;
for(int i=0;i<m&&ste<n-t-1;i++)
{
int u=edge[i].from,v=edge[i].to,val=edge[i].val;
if(des[u]||des[v]) continue;
if(find1(u)!=find1(v))
{
ste++;
ans+=val;
merge1(u,v);
}
}
if(ste==n-t-1) return ans;
return -1;
}
int main()
{
int tes,i,j,res;
scanf("%d",&tes);
while(tes--)
{
int n1,m1,u,v,val,n2,m2;
scanf("%d%d",&n1,&m1);
memset(des,0,sizeof(des));
for(i=0;i<m1;i++)
{
scanf("%d%d%d",&u,&v,&val);
edge[i].from=u;
edge[i].to=v;
edge[i].val=val;
}
scanf("%d%d",&n2,&m2);
n=n1+n2;
m=m1+m2;
for(i=m1;i<m;i++)
{
scanf("%d%d%d",&u,&v,&val);
edge[i].from=u;
edge[i].to=v;
edge[i].val=val;
}
int x;
scanf("%d",&t);
for(i=0;i<t;i++)
{
scanf("%d",&x);
des[x]=1;
}
init();
res=Kruskal();
//cout<<n<<" :n"<<endl;
if(res==-1) puts("what a pity!");
else printf("%d
",res);
}
return 0;
}