ZOJ 3261 Connections in Galaxy War 并查集

来源：互联网发布：西恩潘知乎编辑：程序博客网时间：2024/06/06 00:29

将操作序列反向，用并查集维护集合里最强的星球。

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const int maxn=11000;const int maxm=22000;const int maxq=55000;int n,m,Q;int p[maxn];int a[maxn];int id[maxn];int pa[maxn];void makeset(int n){    for (int i=0;i<=n;i++) pa[i]=i;}int findset(int x){    if (x!=pa[x]) pa[x]=findset(pa[x]);    return pa[x];}void unionset(int x,int y){    x=findset(x);    y=findset(y);    if (x!=y) {        pa[x]=y;        if (a[x]>a[y]){            a[y]=a[x];            id[y]=id[x];        }        else if (a[y]==a[x]&&id[x]<id[y]){            id[y]=id[x];        }    }}struct Dat{    int tp,x,y;};Dat q[maxq];Dat e[maxm];vector<int>ans;set< PII >seto;char s[10];int main(){    int cas=0;    while (~scanf("%d",&n)){        if (cas) printf("\n");        cas++;        rst(q);        rst(e);        ans.clear();        seto.clear();        makeset(n);        for (int i=0;i<n;i++){            scanf("%d",&p[i]);            a[i]=p[i];            id[i]=i;        }        scanf("%d",&m);        for (int i=0;i<m;i++) scanf("%d%d",&e[i].x,&e[i].y);        scanf("%d",&Q);        for (int i=0;i<Q;i++){            scanf("%s",s);            if (s[0]=='d'){                q[i].tp=0;                scanf("%d%d",&q[i].x,&q[i].y);                seto.insert(make_pair(q[i].x,q[i].y));                seto.insert(make_pair(q[i].y,q[i].x));            }            else if (s[0]=='q'){                q[i].tp=1;                scanf("%d",&q[i].x);            }        }        for (int i=0;i<m;i++){            if (seto.find(make_pair(e[i].x,e[i].y))==seto.end()){                unionset(e[i].x,e[i].y);            }        }        for (int i=Q-1;i>=0;i--){            if (q[i].tp){                int fa=findset(q[i].x);                if (a[fa]>p[q[i].x]) ans.push_back(id[fa]);// printf("%d\n",id[fa]);                else ans.push_back(-1);// printf("-1\n");            }            else{                unionset(q[i].x,q[i].y);            }        }        for (int i=sz(ans)-1;i>=0;i--){            printf("%d\n",ans[i]);        }        //printf("\n");    }return 0;}

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