传送门
还是畅通工程
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41447 Accepted Submission(s): 18920
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
prim()
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn = 105;
const int INF = 0x3f3f3f3f;
struct Edge{int u,v,w,next;bool operator < (const Edge &x)const {return w > x.w;}
}edge[maxn*(maxn-1)];
int tot = 0,head[maxn],dis[maxn];
bool vis[maxn];void init()
{tot = 0;memset(head,-1,sizeof(head));
} void addedge(int u,int v,int w)
{edge[tot] = (Edge){u,v,w,head[u]};head[u] = tot++;
}int prim()
{int sum = 0;priority_queue<Edge>que;Edge p;memset(dis,INF,sizeof(dis));memset(vis,false,sizeof(vis));p.v = 1;que.push(p);dis[1] = 0;while (!que.empty()){p = que.top();que.pop();int u = p.v;if (vis[u]) continue;vis[u] = true;sum += dis[u];for (int i = head[u];~i;i = edge[i].next){int v = edge[i].v,w = edge[i].w;if (dis[v] > w){dis[v] = edge[i].w;p.u = u,p.v = v,p.w = w;que.push(p);}}}return sum;
}int main()
{//freopen("input.txt","r",stdin);int N;while (~scanf("%d",&N) && N){int u,v,w;int len = N*(N - 1)/2;init();for (int i = 0;i < len;i++){scanf("%d%d%d",&u,&v,&w);addedge(u,v,w);addedge(v,u,w);}printf("%d\n",prim());}return 0;
}
Kruskal
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef __int64 LL;
const int maxn = 105;
struct Edge{int u,v,next;LL w;
}edge[maxn*(maxn-1)<<1];
int tot = 0,head[maxn],fa[maxn];void init(int N)
{tot = 0;memset(head,-1,sizeof(head));for (int i = 0;i <= N;i++) fa[i] = i;
}bool cmp(struct Edge x,struct Edge y)
{return x.w < y.w;
}int find(int x)
{int r = x;while (r != fa[r]) r = fa[r];int i = x,j;while (i != r){j = fa[i];fa[i] = r;i = j;}return r;
}int main()
{//freopen("input.txt","r",stdin);int N;while (~scanf("%d",&N) && N){LL sum = 0;int len = N*(N-1)/2;init(N);for (int i = 0;i < len;i++){scanf("%d%d%I64d",&edge[i].u,&edge[i].v,&edge[i].w);}sort(edge,edge+len,cmp);for (int i = 0;i < len;i++){int fx = find(edge[i].u),fy = find(edge[i].v);if (fx != fy){fa[fx] = fy;sum += edge[i].w;}}printf("%I64d\n",sum);}return 0;
}
)
)
