POJ 1789&&2485&&1258&&3026

　　这个真的太水了——MST专辑。

　　如果不会MST的两种算法的同学可以出门右转了。

　　大致讲一下，第一题我是用Prim+堆优化的（毕竟点比较多），后面三题用的是Kruskal（习惯打，而且并查集常数实在小）

　　前三题是裸题，最后一题要BFS预处理图上两点间的最短距离再跑Kruskal，稍微麻烦了点

　　按顺序贴自己看吧（第四题数据有坑，数组要开大）

　　1789CODE

#include<cstdio>
#include<string>
#include<queue>
#include<iostream>
#include<cstring>
using namespace std;
const int N=;
struct data
{
    int x,num;
    bool operator <(const data s) const
    {
        return s.x<x;
    }
};
struct edge
{
    int to,next,v;
}e[N*N];
priority_queue <data> tree;
int head[N],dis[N],n,i,j,k,ans;
string s[N];
bool vis[N];
inline int calc(int x,int y)
{
    int res=;
    for (int i=;i<s[x].size();++i)
    if (s[x][i]!=s[y][i]) res++;
    return res;
}
inline void add(int x,int y,int z)
{
    e[++k].to=y; e[k].v=z; e[k].next=head[x]; head[x]=k;
}
int main()
{
    for (;;)
    {
        memset(e,-,sizeof(e));
        memset(head,-,sizeof(head));
        memset(vis,,sizeof(vis));
        memset(dis,,sizeof(dis));
        cin>>n; k=ans=;
        if (!n) break;
        for (i=;i<=n;++i)
        cin>>s[i];
        for (i=;i<n;++i)
        for (j=i+;j<=n;++j)
        {
            int res=calc(i,j);
            add(i,j,res); add(j,i,res);
        }
        dis[]=;
        data p;
        p.x=; p.num=;
        tree.push(p);
        while (!tree.empty())
        {
            int now=tree.top().num;
            tree.pop();
            if (vis[now]) continue;
            vis[now]=; ans+=dis[now];
            for (i=head[now];i!=-;i=e[i].next)
            if (e[i].v<dis[e[i].to])
            {
                dis[e[i].to]=e[i].v;
                data p;
                p.x=dis[e[i].to]; p.num=e[i].to;
                tree.push(p);
            }

        }
        printf("The highest possible quality is 1/%d.\n",ans);
    }
    return ;
}

　　2485CODE

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=;
struct data
{
    int x,y,s;
}a[N*N];
int t,n,i,j,k,x,father[N],ans;
inline void read(int &x)
{
    x=; char ch=getchar();
    while (ch<''||ch>'') ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
}
inline void write(int x)
{
    if (x/) write(x/);
    putchar(x%+'');
}
inline int comp(data a,data b)
{
    return a.s<b.s;
}
inline int getfather(int k)
{
    return father[k]==k?k:father[k]=getfather(father[k]);
}
int main()
{
    read(t);
    while (t--)
    {
        read(n);
        ans=k=;
        for (i=;i<=n;++i)
        for (j=;j<=n;++j)
        {
            read(x);
            if (i==j) continue;
            a[++k].x=i; a[k].y=j; a[k].s=x;
        }
        sort(a+,a+k+,comp);
        for (i=;i<=n;++i)
        father[i]=i;
        for (i=;i<=k;++i)
        {
            int fx=getfather(a[i].x),fy=getfather(a[i].y);
            if (fx!=fy)
            {
                ans=a[i].s;
                father[fx]=fy;
            }
        }
        write(ans); putchar('\n');
    }
    return ;
}

　　

　　1258CODE

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=;
struct data
{
    int x,y,s;
}a[N*N];
int n,i,j,k,x,father[N],ans;
inline void read(int &x)
{
    x=; char ch=getchar();
    while (ch<''||ch>'') ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
}
inline void write(int x)
{
    if (x/) write(x/);
    putchar(x%+'');
}
inline int comp(data a,data b)
{
    return a.s<b.s;
}
inline int getfather(int k)
{
    return father[k]==k?k:father[k]=getfather(father[k]);
}
int main()
{
    while (scanf("%d",&n)!=EOF)
    {
        ans=k=;
        for (i=;i<=n;++i)
        for (j=;j<=n;++j)
        {
            read(x);
            if (i==j) continue;
            a[++k].x=i; a[k].y=j; a[k].s=x;
        }
        sort(a+,a+k+,comp);
        for (i=;i<=n;++i)
        father[i]=i;
        for (i=;i<=k;++i)
        {
            int fx=getfather(a[i].x),fy=getfather(a[i].y);
            if (fx!=fy)
            {
                ans+=a[i].s;
                father[fx]=fy;
            }
        }
        write(ans); putchar('\n');
     }
    return ;
}

　　3026CODE

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=,fx[]={,,-,},fy[]={,,,-};
struct data
{
    int x,y,s;
}e[N*N];
int i,j,n,m,t,k,q[N*N*][],step[N][N],father[N*N],ans;
bool vis[N][N];
char a[N][N];
inline void BFS(int x,int y)
{
    int head=,tail=;
    memset(vis,,sizeof(vis));
    q[][]=x; q[][]=y; vis[x][y]=; step[x][y]=;
    while (head<tail)
    {
        int xx=q[++head][],yy=q[head][];
        if (a[xx][yy]=='S'||a[xx][yy]=='A') e[++k].x=(x-)*m+y,e[k].y=(xx-)*m+yy,e[k].s=step[xx][yy];
        for (int i=;i<;++i)
        {
            int xxx=xx+fx[i],yyy=yy+fy[i];
            if (!vis[xxx][yyy]&&a[xxx][yyy]!='#'&&xxx>&&yyy>&&xxx<=n&&yyy<=m)
            {
                vis[xxx][yyy]=;
                step[xxx][yyy]=step[xx][yy]+;
                q[++tail][]=xxx; q[tail][]=yyy;
            }
        }
    }
}
inline int comp(data a,data b)
{
    return a.s<b.s;
}
inline int getfather(int k)
{
    return father[k]==k?k:father[k]=getfather(father[k]);
}
int main()
{
    scanf("%d\n",&t);
    while (t--)
    {
        scanf("%d%d\n",&m,&n);
        for (k=,ans=,i=;i<=n*m;++i)
        father[i]=i;
        for (i=;i<=n;++i)
        {
            for (j=;j<=m;++j)
            a[i][j]=getchar();
            getchar();
        }
        for (i=;i<=n;++i)
        for (j=;j<=m;++j)
        if (a[i][j]=='S'||a[i][j]=='A') BFS(i,j);
        sort(e+,e+k+,comp);
        for (i=;i<=k;++i)
        {
            int fax=getfather(e[i].x),fay=getfather(e[i].y);
            if (fax!=fay)
            {
                ans+=e[i].s;
                father[fax]=fay;
            }
        }
        printf("%d\n",ans);
    }
}