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Time Limit: 20 Sec Memory Limit: 256 MB
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2966
Input
Standard input will contain multiple test cases. The first line of the input is a single integer T (1 <= T <= 50) which is the number of test cases. And it will be followed by T consecutive test cases.
In each test case, the first line contains two positive integers N and E (2 <= N <= 500, N <= E <= N * (N - 1) / 2), representing the number of the villages and the number of the original power lines between villages. There follow E lines, and each of them contains three integers, A, B, K (0 <= A, B < N, 0 <= K < 1000). A and B respectively means the index of the starting village and ending village of the power line. If K is 0, it means this line still works fine after the snow storm. If K is a positive integer, it means this line will cost K to reconstruct. There will be at most one line between any two villages, and there will not be any line from one village to itself.
Output
For each test case in the input, there‘s only one line that contains the minimum cost to recover the electric system to make sure that there‘s at least one way between every two villages.
Sample Input
1 3 3 0 1 5 0 2 0 1 2 9
5
题意
给你个无向边带权图，让你输出最小生成树
题解：
Kruskal or prim
代码：
//qscqesze #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define test freopen("test.txt","r",stdin) #define maxn 200001 #define mod 10007 #define eps 1e-9 int Num; char CH[20]; //const int inf=0x7fffffff; //нчоч╢С const int inf=0x3f3f3f3f; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } inline void P(int x) { Num=0;if(!x){putchar(‘0‘);puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts(""); } //************************************************************************************** int u[maxn],v[maxn],w[maxn],r[maxn]; int pa[maxn],n,m; bool cmp(int i,int j) { return w[i]<w[j]; } int fi(int x) { return pa[x]==x?x:pa[x]=fi(pa[x]); } int kruskal() { int ans=0; for(int i=0;i<n;i++) pa[i]=i; for(int i=0;i<m;i++) r[i]=i; sort(r,r+m,cmp); for(int i=0;i<m;i++) { int e=r[i]; int x=fi(u[e]); int y=fi(v[e]); if(x!=y) ans+=w[e],pa[x]=y; } return ans; } int main() { //test; int t=read(); while(t--) { n=read(),m=read(); for(int i=0;i<m;i++) { int a=read(),b=read(),c=read(); u[i]=a,v[i]=b,w[i]=c; } printf("%d\n",kruskal()); } }
zoj 2966 Build The Electric System 最小生成树
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原文地址：http://www.cnblogs.com/qscqesze/p/4540149.html
