HDU5692(线段树+dfs序）

Snacks

Time Limit:5000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

百度科技园内有

Input

输入数据第一行是一个整数

Output

对于每组数据，首先输出一行”Case #?:”，在问号处应填入当前数据的组数，组数从1开始计算。

对于每次询问，输出从编号为0的零食机出发，必须经过编号为

Sample Input

1
6 5
0 1
1 2
0 3
3 4
5 3
7 -5 100 20 -5 -7
1 1
1 3
0 2 -1
1 1
1 5

Sample Output

Case #1:
102
27
2
20

线段树＋dfs序

  1 //2016.8.11
  2 #pragma comment(linker, "/STACK:1024000000, 1024000000")
  3 #include<iostream>
  4 #include<cstdio>
  5 #include<vector>
  6 #include<algorithm>
  7 #define N 100005
  8 #define lson (id<<1)
  9 #define rson ((id<<1)|1)
 10 #define mid ((l+r)>>1)
 11 
 12 using namespace std;
 13 
 14 struct node
 15 {
 16     long long max, lazy;
 17 }tree[N*5];
 18 
 19 vector<int> v[N];
 20 long long w[N], val[N];//val[i]表示从根节点走到i节点的价值和
 21 int in[N], out[N], flag;//in记录查询区间的左端，out记录查询区间的右端
 22 
 23 
 24 //线段树模板
 25 //************************************************************************************
 26 void push_up(int id)
 27 {
 28     tree[id].max = max(tree[lson].max, tree[rson].max);
 29     return;
 30 }
 31 
 32 void build_tree(int id, int l, int r)
 33 {
 34     if(l == r)
 35     {
 36         tree[id].max = val[l];
 37         tree[id].lazy = 0;
 38         return ;
 39     }
 40     build_tree(lson, l, mid);
 41     build_tree(rson, mid+1, r);
 42     push_up(id);
 43     tree[id].lazy = 0;
 44     return ;
 45 }
 46 
 47 void push_down(int id, int l, int r)
 48 {
 49     if(tree[id].lazy)
 50     {
 51         tree[lson].lazy += tree[id].lazy;
 52         tree[rson].lazy += tree[id].lazy;
 53         tree[lson].max += tree[id].lazy;
 54         tree[rson].max += tree[id].lazy;
 55         tree[id].lazy = 0;
 56     }
 57     return ;
 58 }
 59 
 60 void ins(int id, int l, int r, int ql, int qr, int tt)
 61 {
 62     if(ql<=l&&r<=qr)
 63     {
 64         tree[id].max += tt;
 65         tree[id].lazy += tt;
 66         return;
 67     }
 68     if(tree[id].lazy) push_down(id, l, r);
 69     if(ql<=mid)
 70           ins(lson, l, mid, ql, qr, tt);
 71     if(mid+1<=qr)
 72           ins(rson, mid+1, r, ql, qr, tt);
 73     push_up(id);
 74     return;
 75 }
 76 
 77 long long query(int id, int l, int r, int ql, int qr)
 78 {
 79     if(ql<=l && r<=qr)
 80     {
 81         return tree[id].max;
 82     }
 83     if(tree[id].lazy)
 84           push_down(id, l, r);
 85     long long maxnum = -9999999999999;
 86     if(ql<=mid) maxnum = max(maxnum, query(lson, l, mid, ql, qr));
 87     if(mid+1<=qr) maxnum = max(maxnum, query(rson, mid+1, r, ql, qr));
 88     return maxnum;
 89 }
 90 //************************************************************************************
 91 
 92 void dfs(int a, int fa, long long wight)
 93 {
 94     in[a] = ++flag;
 95     for(int i = 0; i < v[a].size(); i++)
 96     {
 97         int b = v[a][i];
 98         if(b==fa)continue;
 99         dfs(b, a, w[b]+wight);
100     }
101     out[a] = flag;
102     val[in[a]] = wight;
103 }
104 
105 int main()
106 {
107     int T, n, m, a, b, cmd;
108     cin>>T;
109     for(int kase = 1; kase <= T; kase++)
110     {
111         printf("Case #%d:
", kase);
112         cin>>n>>m;
113         for(int i = 0; i < N; i++)
114             v[i].clear();
115         for(int i = 0; i < n-1; i++)
116         {
117             scanf("%d%d", &a, &b);
118             a++; b++;
119             v[a].push_back(b);
120             v[b].push_back(a);
121         }
122         for(int i = 1 ; i <= n; i++)
123           scanf("%lld", &w[i]);
124         flag = 0;
125         dfs(1, 0, w[1]);
126         for(int i = 1; i <= n; i++)
127           cout<<in[i]<<" "<<out[i]<<endl;
128         build_tree(1, 1, n);
129         while(m--)
130         {
131             scanf("%d", &cmd);
132             if(cmd == 0)
133             {
134                 scanf("%d%d", &a, &b);a++;
135                 ins(1, 1, n, in[a], out[a], b-w[a]);
136                 w[a] = b;
137             }else
138             {
139                 scanf("%d", &a); a++;
140                 long long ans = query(1, 1, n, in[a], out[a]);
141                 cout<<ans<<endl;
142             }
143         }
144     }
145 
146     return 0;
147 }