HDU 1863 畅通工程

来源：互联网发布：linux能运行在arm上吗编辑：程序博客网时间：2024/05/18 00:56

水最小生成树测试下模板直接贴代码了

prim算法的

#include <iostream>#include <cstring>#include <string>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <queue>#include <map>#include<set>#include<stack>#define bug puts("bugbugbug");using namespace std;typedef long long ll;//适合稠密图 复杂度n*n n为点数const int INF=1000000010;const int N=105;//N 为点的数量int vis[N],dis[N];vector<int> V[N];//邻接表存边int g[N][N]={0};//邻接矩阵存边int n,m;int prim() {    memset(vis, 0,sizeof(vis));    for (int i = 1; i <= n; i++) dis[i] = i==1? 0 : INF;//初始化    int ans = 0;    int i;    for (i = 1; i <= n; i++) {        int k = -1, minn = INF;        for (int j = 1; j <= n; j++)            if (!vis[j] && dis[j] < minn) {            minn = dis[j];            k = j;        }        if (k == -1) break;        vis[k] = 1, ans += dis[k];        for (int j = 0; j < V[k].size(); j++) {            int u = V[k][j];            if (!vis[u] && g[k][u] < dis[u]) dis[u] = g[k][u];        }    }    if(i!=n+1)return -1;    return ans;}int main(){    while(~scanf("%d%d",&m,&n)&&m)    {        for(int i=0;i<=n;i++)        V[i].clear();        memset(g,0,sizeof(g));        for(int i=0;i<m;i++){                int a,b,c;            scanf("%d%d%d",&a,&b,&c);        V[a].push_back(b);         V[b].push_back(a);         g[a][b]=g[b][a]=c;        }             int ans=prim();             if(ans==-1) puts("?");             else printf("%d\n",ans);    }}

Kruscal的

#include <iostream>#include <cstring>#include <string>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <queue>#include <map>#include<set>#include<stack>#define bug puts("bugbugbug");using namespace std;typedef long long ll;const int maxm=10005;int f[105];int n,m;struct node{    int x, y, v;    bool operator<(const node &b)const{        if(this->v!=b.v)    return (this->v<b.v);    if(this->x!=b.x)    return (this->x<b.x);    return (this->y<b.y);    }} e[maxm];int find(int x){ return f[x]= f[x]==x?x:find(f[x]); }int Kruscal(){    sort(e, e + m);    int cnt = n, ans = 0;    for(int i = 0; i < n; i++) f[i] = i;    for (int i = 0; i < m; i++)    {        int x = find(e[i].x);        int y = find(e[i].y);        if (x != y)        {            f[x] = y;            cnt--;            ans += e[i].v;            if (cnt == 1) break;        }    }    if(cnt!=1)return -1;    return ans;}int main(){    while(~scanf("%d%d",&m,&n)&&m)    {        for(int i=0;i<m;i++)            scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].v);             int ans=Kruscal();             if(ans==-1) puts("?");             else printf("%d\n",ans);    }}

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