bzoj1455: 罗马游戏

来源：互联网发布：网络主播直播内容策划编辑：程序博客网时间：2024/06/09 17:10

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=1455

思路：左偏树练习题

用并查集维护连通，然后开个数组记录每个人是否已被杀死，用可并堆支持合并和求最小值

左偏树是一种支持合并的堆，写起来比手写堆还要短...

只有一个操作，merge(a,b)，就是把a，b合并...

具体构建参见论文：http://wenku.baidu.com/link?url=CPzCT6_74LbTMczlGZL8oB3wa9vivG2U9pTEXFMDLuolIvRgp_cbi2BHwT3NXhher9qgjXWEoB8PG3P8QeT2A22Zv3L5oi7nSADoPG08ofa

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>const int maxn=1000010;using namespace std;int n,m,fa[maxn];bool die[maxn];char op[2];int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}void read(int &x){char ch;for (ch=getchar();!isdigit(ch);ch=getchar());for (x=0;isdigit(ch);ch=getchar()) x=x*10+ch-'0';}struct Ltree{int tot,l[maxn],r[maxn],dis[maxn],v[maxn];void newnode(int val){v[++tot]=val,l[tot]=r[tot]=dis[tot]=0;}int merge(int x,int y){if (!x) return y;if (!y) return x;if (v[x]>v[y]) swap(x,y);r[x]=merge(r[x],y);if (dis[r[x]]>dis[l[x]]) swap(l[x],r[x]);dis[x]=dis[r[x]]+1;return x;}}h;int main(){scanf("%d",&n);for (int i=1,x;i<=n;i++) read(x),h.newnode(x),fa[i]=i;scanf("%d",&m);for (int i=1,x,y;i<=m;i++){scanf("%s",op);if (op[0]=='M'){scanf("%d%d",&x,&y);if (die[x]||die[y]) continue;x=find(x),y=find(y);if (x!=y) fa[x]=fa[y]=h.merge(x,y);}else{scanf("%d",&x);if (die[x]){puts("0");continue;}else{x=find(x),die[x]=1;printf("%d\n",h.v[x]);fa[x]=h.merge(h.l[x],h.r[x]);fa[fa[x]]=fa[x];}}}return 0;}

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