2012 Multi-University Training Contest 7
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2012 Multi-University Training Contest 7
A.As long as Binbin loves Sangsang
题意:给你一棵树,每个结点有两个属性值,1.能力值,2.忠诚度。然后m个询问,每次询问一个整数u,求u的子树中能力值大于u的且忠诚度最大的点的编号。
SOL:先按能力值排序,这样从大到小考虑就满足了条件1,然后从大到小依次在线段树里查询子树中忠诚度最大的点的编号,复杂度O(nlogn)。
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #define tl (p << 1) 6 #define tr (p << 1 | 1) 7 using namespace std; 8 const int LEN = 1e5 + 5; 9 int i, j, k, n, m, s, t, tot, Time; 10 struct edge { 11 int vet, next; 12 } E[LEN * 2]; 13 struct node { 14 int x, y, id; 15 } a[LEN]; 16 bool cmp(const node &x, const node &y) { 17 return x.x > y.x; 18 }; 19 int head[LEN], size[LEN], tid[LEN], ans[LEN], pre[LEN], b[LEN]; 20 int to[1000005]; 21 int tmax[LEN * 4]; 22 void add(int u, int v) { 23 E[++tot] = (edge){v, head[u]}; 24 head[u] = tot; 25 } 26 void dfs(int u) { 27 size[u] = 1; 28 tid[u] = ++Time; 29 pre[Time] = u; 30 for (int e = head[u]; e != -1; e = E[e].next) { 31 int v = E[e].vet; 32 dfs(v); 33 size[u] += size[v]; 34 } 35 } 36 int ask(int l, int r, int x, int y, int p) { 37 if (l == x && y == r) { 38 return tmax[p]; 39 } 40 int mid = (l + r) >> 1; 41 if (mid >= y) { 42 return ask(l, mid, x, y, tl); 43 } else if (mid + 1 <= x) { 44 return ask(mid + 1, r, x, y, tr); 45 } else { 46 return max(ask(l, mid, x, mid, tl), ask(mid + 1, r, mid + 1, y, tr)); 47 } 48 } 49 void update(int p) { 50 tmax[p] = max(tmax[tl], tmax[tr]); 51 } 52 void modify(int l, int r, const int &x, int p, const int &c) { 53 if (l == r) { 54 tmax[p] = c; 55 return; 56 } 57 int mid = (l + r) >> 1; 58 if (mid >= x) { 59 modify(l, mid, x, tl, c); 60 } else { 61 modify(mid + 1, r, x, tr, c); 62 } 63 update(p); 64 } 65 void build(int l, int r, int p) { 66 if (l == r) { 67 tmax[p] = -1; 68 return; 69 } 70 int mid = (l + r) >> 1; 71 build(l, mid, tl); 72 build(mid + 1, r, tr); 73 update(p); 74 } 75 int main() { 76 int T; 77 scanf("%d", &T); 78 while (T--) { 79 tot = Time = 0; 80 scanf("%d %d", &n, &m); 81 for (int i = 1; i <= n; i++) { 82 head[i] = -1; 83 size[i] = 0; 84 tid[i] = 0; 85 } 86 a[1] = (node){1e9, 0, 1}; 87 for (int i = 2; i <= n; i++) { 88 int fa, x, y; 89 scanf("%d %d %d", &fa, &x, &y); 90 x++,y++,fa++; 91 swap(x, y); 92 add(fa, i); 93 a[i] = (node){x, y, i}; 94 to[y] = i; 95 } 96 dfs(1); 97 build(1, n, 1); 98 sort(a + 1, a + 1 + n, cmp); 99 int j = 1; 100 for (int i = 2; i <= n; i++) { 101 int x = a[i].id, t = ask(1, n, tid[x], tid[x] + size[x] - 1, 1); 102 if (t == -1) { 103 ans[x] = 0; 104 } else { 105 ans[x] = to[t]; 106 } 107 while (j + 1 <= i && a[j + 1].x > a[i + 1].x) { 108 modify(1, n, tid[a[j + 1].id], 1, a[j + 1].y); 109 j++; 110 } 111 } 112 while (m--) { 113 int x; 114 scanf("%d", &x); 115 printf("%d ", ans[x + 1] - 1); 116 } 117 } 118 return 0; 119 }
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