hdu 1512 Monkey King

并查集显然要用到

左偏树解决合并的问题

 

这题体现了数据结构的强大啊！

 

#include<stdio.h> #include <algorithm> using namespace std; const int MAXN = 100100; int father[MAXN]; struct Monkey{ int l,r; int dis; int strong; }LTree[MAXN]; int find(int x){ if(x != father[x]) father[x] = find(father[x]); return father[x]; } int merge(int x,int y){ //返回合并后的根 if( x == 0 ) return y; if( y == 0) return x; if( LTree[x].strong < LTree[y].strong) //大顶堆 swap(x,y); LTree[x].r = merge(LTree[x].r,y); //递归合并右子树和Y int l = LTree[x].l , r = LTree[x].r; father[r] = x; //更新T右子树的根 if(LTree[l].dis < LTree[r].dis) //维护堆性质 swap(LTree[x].l,LTree[x].r); if(LTree[x].r == 0) //如果没有右子树 则距离为0 LTree[x].dis = 0; else LTree[x].dis = LTree[LTree[x].r].dis + 1; return x; } int del(int x){ //返回删除根以后左右子树的合并的根 int l,r; l = LTree[x].l; r = LTree[x].r; father[l] = l; father[r] = r; LTree[x].l = LTree[x].r = LTree[x].dis = 0; return merge(l,r); } void solve(int x,int y){ LTree[x].strong /= 2; LTree[y].strong /= 2; //问每次PK以后，当前这个群体里力量最大的猴子的力量是多少。 int left,right; left = del(x); right = del(y); left = merge(left,x); right = merge(right,y); left = merge(left,right); printf("%d/n",LTree[left].strong); } int main(){ int n,m,i,x,y; while(scanf("%d",&n) == 1){ for(i = 1; i <= n; i++){ scanf("%d",<ree[i].strong); LTree[i].l = 0; LTree[i].r = 0; LTree[i].dis = 0; father[i] = i; //起始已自己为父亲 } scanf("%d",&m); for(i = 1; i <= m; i++){ scanf("%d%d",&x,&y); int fx = find(x),fy = find(y); if(fx == fy){ printf("-1/n"); } else { solve(fx,fy); } } } return 0; }