还是畅通工程
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29254 Accepted Submission(s): 13088
Problem Description
某省调查乡村交通状况,得到的统计表中列出了任意两村庄间的距离。省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通(但不一定有直接的公路相连,只要能间接通过公路可达即可),并要求铺设的公路总长度为最小。请计算最小的公路总长度。
Input
测试输入包含若干测试用例。每个测试用例的第1行给出村庄数目N ( < 100 );随后的N(N-1)/2行对应村庄间的距离,每行给出一对正整数,分别是两个村庄的编号,以及此两村庄间的距离。为简单起见,村庄从1到N编号。
当N为0时,输入结束,该用例不被处理。
Output
对每个测试用例,在1行里输出最小的公路总长度。
Sample Input
3
1 2 1
1 3 2
2 3 4
4
1 2 1
1 3 4
1 4 1
2 3 3
2 4 2
3 4 5
0
Sample Output
3
5
Hint
Hint
Huge input, scanf is recommended.
Source
浙大计算机研究生复试上机考试-2006年
prim 算法
//#include<bits/stdc++.h> #include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<string> #include<queue> #include<vector> #include<map> #include<cstdlib> #include<set> #include<stack> #include<cstring> using namespace std; template<class T>inline T read(T&x) { char c; while((c=getchar())<=32)if(c==EOF)return -1; bool ok=false; if(c=='-')ok=true,c=getchar(); for(x=0; c>32; c=getchar()) x=x*10+c-'0'; if(ok)x=-x; return 1; } template<class T> inline T read_(T&x,T&y) { return read(x)!=-1&&read(y)!=-1; } template<class T> inline T read__(T&x,T&y,T&z) { return read(x)!=-1&&read(y)!=-1&&read(z)!=-1; } template<class T> inline void write(T x) { if(x<0)putchar('-'),x=-x; if(x<10)putchar(x+'0'); else write(x/10),putchar(x%10+'0'); } template<class T>inline void writeln(T x) { write(x); putchar('\n'); } //-------ZCC IO template------ const int maxn=101; const double inf=999999999; #define lson (rt<<1),L,M #define rson (rt<<1|1),M+1,R #define M ((L+R)>>1) #define For(i,t,n) for(int i=(t);i<(n);i++) typedef long long LL; typedef double DB; typedef pair<int,int> P; #define bug printf("---\n"); #define mod 10007 int G[maxn][maxn]; int vis[maxn]; int mincost[maxn]; int V,E; void prim() { For(i,0,V+1) { mincost[i]=inf; vis[i]=false; } mincost[1]=0; int ans=0; while(true) { int v=-1; For(u,1,V+1)if(!vis[u]&&(v==-1||mincost[v]>mincost[u]))v=u; if(v==-1)break; vis[v]=true; ans+=mincost[v]; For(u,1,V+1)mincost[u]=min(mincost[u],G[v][u]); } writeln(ans); } int main() { //#ifndef ONLINE_JUDGE //freopen("in.txt","r",stdin); // #endif // ONLINE_JUDGE int n,m,i,j,k,t; while(read(V)&&V) { E=V*(V-1)/2; while(E--) { int a,b,c; read__(a,b,c); G[a][b]=G[b][a]=c; } prim(); } return 0; }
Kruskal算法
//#include<bits/stdc++.h> #include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<string> #include<queue> #include<vector> #include<map> #include<cstdlib> #include<set> #include<stack> #include<cstring> using namespace std; template<class T>inline T read(T&x) { char c; while((c=getchar())<=32)if(c==EOF)return -1; bool ok=false; if(c=='-')ok=true,c=getchar(); for(x=0; c>32; c=getchar()) x=x*10+c-'0'; if(ok)x=-x; return 1; } template<class T> inline T read_(T&x,T&y) { return read(x)!=-1&&read(y)!=-1; } template<class T> inline T read__(T&x,T&y,T&z) { return read(x)!=-1&&read(y)!=-1&&read(z)!=-1; } template<class T> inline void write(T x) { if(x<0)putchar('-'),x=-x; if(x<10)putchar(x+'0'); else write(x/10),putchar(x%10+'0'); } template<class T>inline void writeln(T x) { write(x); putchar('\n'); } //-------ZCC IO template------ const int maxn=101; const double inf=999999999; #define lson (rt<<1),L,M #define rson (rt<<1|1),M+1,R #define M ((L+R)>>1) #define For(i,t,n) for(int i=(t);i<(n);i++) typedef long long LL; typedef double DB; typedef pair<int,int> P; #define bug printf("---\n"); #define mod 10007 struct edge { int from,to,cost; edge(){} edge(int a,int b,int c){from=a,to=b,cost=c;} bool operator<(const edge&tmp)const{ return cost<tmp.cost; } }es[maxn*maxn]; int f[maxn]; void init(int n) { For(i,0,n+1)f[i]=i; } int find(int x) { return x==f[x]?x:(f[x]=find(f[x])); } void unio(int x,int y) { x=find(x);y=find(y); if(x==y)return ; f[x]=y; } bool same(int x,int y) { return find(x)==find(y); } int main() { //#ifndef ONLINE_JUDGE //freopen("in.txt","r",stdin); // #endif // ONLINE_JUDGE int n,m,i,j,k,t; int V,E; while(read(V)&&V) { E=V*(V-1)/2; for(i=0;i<E;i++) { int a,b,c; read__(a,b,c); es[i]=edge(a,b,c); } init(V); sort(es,es+E); int ans=0; for(i=0;i<E;i++) { edge e=es[i]; if(!same(e.to,e.from)) { ans+=e.cost; unio(e.to,e.from); } } writeln(ans); } return 0; }
时间: 2024-10-09 09:53:44