[破碎的状态] Gym 100960G
感谢@似水流年 翻译!
题意:
给你n个数字
你需要算出它们排序后,有多少个数字>=前面所有数之和
注意:
1 1 1 1 1中,前面两个1是成立的(sum=0/1),后面三个是不成立的(sum=2/3/4)
另外会有M次修改,求修改前和每次修改后的答案
解法:
treap直接上
当然这treap需要记录father了...
insert_node:没区别
delete_node:注意要支持删从某点到根的
rotate:注意sum和father的更改
关键是那个dfs函数,求出某个地方的find_max看是否可行,如果可能有个数大于它们的全部我们就去试试找解
反正解也不多最多大概40~50个吧..
代码:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 | #include<set>#include<map>#include<list>#include<queue>#include<stack>#include<math.h>#include<string>#include<time.h>#include<bitset>#include<vector>#include<memory>#include<utility>#include<stdio.h>#include<sstream>#include<fstream>#include<iostream>#include<stdlib.h>#include<string.h>#include<algorithm>using namespace std;long long a[100005];struct node{ long long val; int weight; long long sum; int size; node * father; node * ch[2];};node * null;node * root;node * b[100005];node * new_node(long long x){ static node a[100005]; static int top=0; node * t=&a[top++]; t->val=x; t->sum=x; t->weight=rand(); t->size=1; t->father=null; t->ch[0]=null; t->ch[1]=null; return t;}void rotate(node * &x,int c){ node * y=x->ch[c]; x->ch[c]=y->ch[!c]; y->ch[!c]->father=x; y->ch[!c]=x; y->sum=x->sum; y->size=x->size; y->father=x->father; x->sum=x->ch[0]->sum+x->ch[1]->sum+x->val; x->size=x->ch[0]->size+x->ch[1]->size+1; x->father=y; x=y;}void insert_node(node * &x,node * y){ if (x==null) { x=y; return; } x->size++; x->sum+=y->sum; int c; if (y->val>x->val) { c=1; } else { c=0; } y->father=x; insert_node(x->ch[c],y); if (x->ch[c]->weight>x->weight) rotate(x,c);}void delete_node(node * &x,long long y){ x->size--; x->sum-=y; if (x->val==y) { int c; if (x->ch[0]->weight>x->ch[1]->weight) { c=0; } else { c=1; } if (x->ch[c]==null) { x=null; return; } rotate(x,c); delete_node(x->ch[!c],y); } else { if (x->val<y) { delete_node(x->ch[1],y); } else { delete_node(x->ch[0],y); } }}int n;long long max_val(node * x){ if (x->ch[1]==null) return x->val; return max_val(x->ch[1]);}int dfs(node * x,long long y){ int sum=0; if (x->ch[0]!=null) { if (max_val(x->ch[0])>=y) { sum+=dfs(x->ch[0],y); } } if (x->val>=y+x->ch[0]->sum) { sum++; } if (x->ch[1]!=null) { if (max_val(x->ch[1])<y+x->ch[0]->sum+x->val) return sum; sum+=dfs(x->ch[1],y+x->ch[0]->sum+x->val); } return sum;}int query(){ return dfs(root,0);}int main(){ #ifdef absi2011 freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); #endif ios::sync_with_stdio(false); cin>>n; int i; null=new_node(0); null->ch[0]=null; null->ch[1]=null; null->father=null; null->weight=-1; null->size=0; root=null; for (i=0;i<n;i++) { cin>>a[i]; b[i]=new_node(a[i]); insert_node(root,b[i]); } printf("%d\n",query()); int m; cin>>m; for (i=0;i<m;i++) { int x; long long y; cin>>x>>y; x--; node * t=b[x]->father; for (;t!=null;) { t->size--; t->sum-=b[x]->val; t=t->father; } if (b[x]==root) { delete_node(root,b[x]->val); } else { int c; if (b[x]->father->ch[0]==b[x]) c=0; else c=1; delete_node(b[x]->father->ch[c],b[x]->val); } b[x]->val=y; b[x]->sum=y; b[x]->size=1; b[x]->ch[0]=null; b[x]->ch[1]=null; b[x]->father=null; insert_node(root,b[x]); printf("%d\n",query()); } return 0;} |
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