Description
The cows, who always have an inferiority complex about their intelligence, have a new guessing game to sharpen their brains.
A designated 'Hay Cow' hides behind the barn and creates N (1 ≤ N ≤ 1,000,000) uniquely-sized stacks (conveniently numbered 1..N) of hay bales, each with 1..1,000,000,000 bales of hay.
The other cows then ask the Hay Cow a series of Q (1 ≤ Q ≤ 25,000) questions about the the stacks, all having the same form:
What is the smallest number of bales of any stack in the range of stack numbers Ql..Qh (1 ≤ Ql ≤ N; Ql ≤ Qh ≤ N)?The Hay Cow answers each of these queries with a single integer A whose truthfulness is not guaranteed.
Help the other cows determine if the answers given by the Hay Cow are self-consistent or if certain answers contradict others.
给一段长度为n，每个位置上的数都不同的序列a[1..n]和q和问答，每个问答是(x, y, r)代表RMQ(a, x, y) = r, 要你给出最早的有矛盾的那个问答的编号。
Input
-
Line 1: Two space-separated integers: N and Q
- Lines 2..Q+1: Each line contains three space-separated integers that represent a single query and its reply: Ql, Qh, and A
Output
- Line 1: Print the single integer 0 if there are no inconsistencies among the replies (i.e., if there exists a valid realization of the hay stacks that agrees with all Q queries). Otherwise, print the index from 1..Q of the earliest query whose answer is inconsistent with the answers to the queries before it.
Sample Input
20 4
1 10 7
5 19 7
3 12 8
11 15 12
Sample Output
3
题解
二分求解。
二分答案，将答案范围内的最小值$Ai$进行降序排序 然后我们可以观察一下得到的这些区间
对于不同$Ai$想一想如果它被之前出现的区间（比它大的$Ai$）都覆盖了，那么肯定就是有矛盾的
给点提示：对于同样的$Ai$询问要用交集，覆盖要用并集
这样就可以很明显地用线段树来搞了
1 //It is made by Awson on 2017.10.27 2 #include <set> 3 #include <map> 4 #include <cmath> 5 #include <ctime> 6 #include <queue> 7 #include <stack> 8 #include <vector> 9 #include <cstdio> 10 #include <string> 11 #include <cstdlib> 12 #include <cstring> 13 #include <iostream> 14 #include <algorithm> 15 #define LL long long 16 #define Max(a, b) ((a) > (b) ? (a) : (b)) 17 #define Min(a, b) ((a) < (b) ? (a) : (b)) 18 #define Lr(o) (o<<1) 19 #define Rr(o) (o<<1|1) 20 using namespace std; 21 const int N = 1000000; 22 const int INF = ~0u>>1; 23 24 int n, q; 25 struct tt { 26 int l, r, a; 27 } a[N+5], b[N+5]; 28 bool comp(const tt &a, const tt &b) { 29 if (a.a != b.a) return a.a > b.a; 30 return a.l == b.l ? a.r < b.r : a.l < b.l; 31 } 32 struct segment { 33 int sgm[(N<<2)+5], lazy[(N<<2)+5]; 34 void build(int o, int l, int r) { 35 lazy[o] = 0; 36 if (l == r) { 37 sgm[o] = INF; return; 38 } 39 int mid = (l+r)>>1; 40 build(Lr(o), l, mid); 41 build(Rr(o), mid+1, r); 42 sgm[o] = Max(sgm[Lr(o)], sgm[Rr(o)]); 43 } 44 void pushdown(int o) { 45 if (lazy[o]) { 46 sgm[Lr(o)] = sgm[Rr(o)] = lazy[Lr(o)] = lazy[Rr(o)] = lazy[o]; 47 lazy[o] = 0; 48 } 49 } 50 void update(int o, int l, int r, int a, int b, int key) { 51 if (a <= l && r <= b) { 52 sgm[o] = lazy[o] = key; return; 53 } 54 pushdown(o); 55 int mid = (l+r)>>1; 56 if (a <= mid) update(Lr(o), l, mid, a, b, key); 57 if (mid < b) update(Rr(o), mid+1, r, a, b, key); 58 sgm[o] = Max(sgm[Lr(o)], sgm[Rr(o)]); 59 } 60 int query(int o, int l, int r, int a, int b) { 61 if (a <= l && r <= b) return sgm[o]; 62 pushdown(o); 63 int mid = (l+r)>>1; 64 int a1 = 0, a2 = 0; 65 if (a <= mid) a1 = query(Lr(o), l, mid, a, b); 66 if (mid < b) a2 = query(Rr(o), mid+1, r, a, b); 67 return Max(a1, a2); 68 } 69 }T; 70 71 bool get(int l, int r, int &x, int &y) { 72 int ll = b[l].l, rr = b[l].r; 73 for (int i = l+1; i <= r; i++) { 74 int lll = b[i].l, rrr = b[i].r; 75 if (rr < lll) return false; 76 ll = lll; 77 } 78 x = ll, y = rr; 79 return true; 80 } 81 bool judge(int mid) { 82 T.build(1, 1, n); 83 for (int i = 1; i <= mid; i++) b[i] = a[i]; 84 sort(b+1, b+1+mid, comp); 85 for (int i = 1; i <= mid; i++) { 86 int loc, l, r; 87 for (loc = i; loc <= mid; loc++) if (b[loc].a != b[i].a) break; 88 loc--; 89 if (!get(i, loc, l, r)) return false; 90 int t = T.query(1, 1, n, l, r); 91 if (t != INF && t != b[i].a) return false; 92 for (int k = i; k <= loc; k++) 93 T.update(1, 1, n, b[k].l, b[k].r, b[k].a); 94 i = loc; 95 } 96 return true; 97 } 98 void work() { 99 scanf("%d%d", &n, &q); 100 for (int i = 1; i <= q; i++) 101 scanf("%d%d%d", &a[i].l, &a[i].r, &a[i].a); 102 int L = 1, R = q, ans = 0; 103 while (L <= R) { 104 int mid = (L+R)>>1; 105 if (judge(mid)) L = mid+1; 106 else R = mid-1, ans = mid; 107 } 108 printf("%d ", ans); 109 } 110 int main() { 111 work(); 112 return 0; 113 }