并查集专题

洛谷P3144 [USACO16OPEN]关闭农场关闭农场
离线的反着的并查集
看看在不在一个集合内

#include<cstdio>#include<queue>#include<cstring>#include<iostream>using namespace std;int n,m;int f[99999],a[3009][3099],b[3999];int ans[19999];int find(int x){    if(x==f[x])return x;    int fx=find(f[x]);    return f[x]=fx; }int main(){    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++)    f[i]=i;    for(int i=1;i<=m;i++)    {        int x,y;        scanf("%d%d",&x,&y);        a[x][y]=1;        a[y][x]=1;    }    for(int i=1;i<=n;i++)    scanf("%d",&b[i]);    for(int i=n;i>=1;i--)    {        int fx=find(b[i]);        for(int j=i;j<=n;j++)        if(a[b[i]][b[j]])        {            int fy=find(b[j]);            f[fy]=fx;        }         int w=0,fz=0;        for(int j=i;j<=n;j++)        {            fz=find(b[j]);            if(w!=fz&&w!=0) ans[i]++;            w=fz;        }    }    for(int i=1;i<=n;i++)    if(ans[i])printf("NO\n");    else printf("YES\n"); }

poj 1984导航噩梦
带权并查集
不过我就是过不了
请大佬给我找错2333
思路：
维护横坐标和纵坐标
通过输入的x-x1，y-y1
判断 父节点与父节点的距离
合并父节点

#include<cstdio>#include<queue>#include<cstring>#include<iostream>using namespace std;int n,m;int f[99999],disx[99999],disy[99999],f1[99999],f2[99999],l[94999];char d[99999];int w[99999],r[99999],s[99999];int abs(int x){    return x>0?x:-x;}int find(int x){    if(x==f[x])return x;    int fx=find(f[x]);    disx[x]+=disx[f[x]];    disy[x]+=disy[f[x]];    return f[x]=fx; }int main(){    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++)    f[i]=i;    for(int i=1;i<=m;i++)        {            scanf("%d%d%d",&f1[i],&f2[i],&l[i]);            cin>>d[i];        }        int t;    scanf("%d",&t);    for(int i=1;i<=t;i++)    {        scanf("%d%d%d",&w[i],&r[i],&s[i]);    }    int j=1;    for(int i=1;i<=m;i++)    {        int fx1=find(f1[i]);        int fx2=find(f2[i]);        if(fx1==fx2) continue;        if(d[i]=='E')            disx[f1[i]]=disx[f1[i]]-disx[f2[i]]+l[i],disy[f1[i]]=disy[f1[i]]-disy[f2[i]];        if(d[i]=='W')            disx[f1[i]]=disx[f1[i]]-disx[f2[i]]-l[i],disy[f1[i]]=disy[f1[i]]-disy[f2[i]];        if(d[i]=='N')            disy[f1[i]]=disy[f1[i]]-disy[f2[i]]+l[i],disx[f1[i]]=disx[f1[i]]-disx[f2[i]];        if(d[i]=='S')            disy[f1[i]]=disy[f1[i]]-disy[f2[i]]-l[i],disx[f1[i]]=disx[f1[i]]-disx[f2[i]];        f[fx1]=fx2;        if(s[j]==i){         fx1=find(w[j]);         fx2=find(r[j]);        if(fx1!=fx2)        {            j++;            printf("-1\n");            continue;        }        printf("%d\n",abs(disx[w[j]]-disx[r[j]])+abs(disy[w[j]]-disy[r[j]]));        j++;        }    }}