线段树(2)

hdu 4366 Successor

做法：对每个人按照ability由大到小排序，把loyalty插入到线段树里面，dfs处理出每个点所在的区间，然后区间查询，单点更新。（这里学到了查询区间最大值所在id的方法）。。

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #include <cmath>
  6 #include <queue>
  7 #include <stack>
  8
  9 using namespace std;
 10
 11 #define LL long long
 12 #define eps 1e-6
 13 #define inf 0x3f3f3f3f
 14 #define MP make_pair
 15 #define N 250020
 16 #define M 1000020
 17 #pragma comment(linker, "/STACK:1024000000,1024000000")
 18 #define Pi acos(-1.0)
 19 #define ls (i << 1)
 20 #define rs (ls | 1)
 21 #define md ((ll + rr) >> 1)
 22 #define lson ll, md, ls
 23 #define rson md + 1, rr, rs
 24
 25 int fst[N], vv[N], nxt[N], e;
 26 void init(){
 27     memset(fst, -1, sizeof fst); e = 0;
 28 }
 29 void add(int u, int v){
 30     vv[e] = v, nxt[e] = fst[u], fst[u] = e++;
 31 }
 32 int in[N], out[N], lab[N], dc;
 33 void dfs(int u){
 34     in[u] = ++dc;
 35     lab[dc] = u;
 36     for(int i = fst[u]; ~i; i = nxt[i]){
 37         int v = vv[i];
 38         dfs(v);
 39     }
 40     out[u] = dc;
 41 }
 42 struct node{
 43     int id, val, loy;
 44     node(){}
 45     node(int _loy, int _val, int _id){
 46         loy = _loy, val = _val, id = _id;
 47     }
 48     bool operator < (const node &b) const{
 49         return val > b.val;
 50     }
 51 }b[N];
 52
 53 int mx[N<<2], val[N];
 54 void build(int ll, int rr, int i){
 55     mx[i] = -1;
 56     if(ll == rr) return ;
 57     build(lson), build(rson);
 58 }
 59 void push_up(int i){
 60     if(val[mx[ls]] > val[mx[rs]])
 61         mx[i] = mx[ls];
 62     else mx[i] = mx[rs];
 63 }
 64 void update(int p, int v, int ll, int rr, int i){
 65     if(ll == rr){
 66         val[ll] = v;
 67         mx[i] = ll;
 68         return ;
 69     }
 70     if(p <= md) update(p, v, lson);
 71     else update(p, v, rson);
 72     push_up(i);
 73 }
 74 int query(int l, int r, int ll, int rr, int i){
 75     if(l > r) return -1;
 76     if(ll == l && r == rr)
 77         return mx[i];
 78     if(r <= md) return query(l, r, lson);
 79     else if(l > md) return query(l, r, rson);
 80     else{
 81         int ret1 = query(l, md, lson);
 82         int ret2 = query(md + 1, r, rson);
 83         if(ret1 == -1) return ret2;
 84         if(ret2 == -1) return ret1;
 85         return val[ret1] >= val[ret2] ? ret1 : ret2;
 86     }
 87 }
 88
 89 int ans[N];
 90 int main(){
 91     int cas;
 92     scanf("%d", &cas);
 93     while(cas--){
 94         int n, m;
 95         scanf("%d%d", &n, &m);
 96         init();
 97         for(int i = 1; i < n; ++i){
 98             int u, val, loy;
 99             scanf("%d%d%d", &u, &loy, &val);
100             add(u, i);
101             b[i] = node(loy, val, i);
102         }
103         dc = 0;
104         dfs(0);
105         sort(b + 1, b + n);
106         build(1, n, 1);
107         memset(val, -1, sizeof val);
108         for(int i = 1, j; i < n; i = j){
109             j = i;
110             while(j < n && b[i].val == b[j].val){
111                 int k = b[j].id;
112                 int t = query(in[k] + 1, out[k], 1, n, 1);
113                 if(t == -1) ans[k] = -1;
114                 else ans[k] = lab[t];
115                 j++;
116             }
117             j = i;
118             while(j < n && b[i].val == b[j].val){
119                 update(in[b[j].id], b[j].loy, 1, n, 1);
120                 j++;
121             }
122         }
123         while(m--){
124             int i;
125             scanf("%d", &i);
126             printf("%d\n", ans[i]);
127         }
128     }
129     return 0;
130 }