POJ3657.Haybale Guessing

http://poj.org/problem?id=3657

Haybale Guessing

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1180 Accepted: 359

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 Q_l..Q_h (1 ≤ Q_l ≤ N; Q_l ≤ Q_h ≤ 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.

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: Q_l, Q_h, 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

Source

USACO 2008 January Gold

题意：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。

大致思路：二分答案qi。然后将前qi个问答按r从大到小排序，用染色的思想——如果当前问答的区间已经被全部染色了，就表示前qi的问答构成有矛盾。

判断染色的方法有很多种，本人一开始想到线段树，后来在别人的指导下学会用并查集（并查集要比线段树好写很多），具体的方法就是把某个问答区间[x, y]所有值的父亲都只想区间头x，关键在于枚举是从后枚举，而且那些祖先不是他自己的可以不用判断，也就是说枚举到一个数，其实可以直接跳到他祖先那里，因为中间肯定都已经染了色。在有并查集的思想只需要吧他祖先的父亲指向新的点就可以了。。这样均摊下来每次只需O(n), 总的时间复杂度为O(nlogm)，离散化可以做到O(mlongm)。还有就是判断是否有未染色区间用含有相同r的区间的交去判断，染色是用这些区间的并染色。

#include <map>#include <cmath>#include <queue>#include <string>#include <vector>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define MAXM 25100#define MAXN 1100000#define eps (1e-8)#define INF 1000000000#define abs(x) ((x) > 0? (x): -(x))#define sqr(x) ((x) * (x))#define MAX(a, b) ((a) > (b)? (a): (b))#define MIN(a, b) ((a) < (b)? (a): (b))typedef long long LL;struct QUERY{    int x, y, a, id;} List[MAXM];int n, m, fa[MAXN];bool colored[MAXN];bool cmp_a(QUERY p, QUERY q){    return p.a > q.a;}bool cmp_id(QUERY p, QUERY q){    return p.id < q.id;}void init(){    for (int i = 1; i <= m; ++i)    {        scanf("%d%d%d", &List[i].x, &List[i].y, &List[i].a);        List[i].id = i;    }}int find(int x){    int y = x;    while (fa[x] != x) x = fa[x];    while (y != x)    {        int tf = fa[y];        fa[y] = x;        y = tf;    }    return x;}bool color(int x, int y){    bool ret = false;    for (int i = y; i >= x; --i)    {        if (fa[i] == i && !colored[i]) ret = true;        if (fa[i] == i) fa[i] = x; else        {            int anc = find(i);            if (anc >= x) fa[anc] = x;            i = anc;        }    }    if (fa[x] == x) colored[x] = true;    return ret;}int main(){    while (scanf("%d%d", &n, &m) != EOF)    {        init();        int L = 1, R = m + 1;        while (L < R)        {            memset(colored, 0, sizeof(colored));            for (int i = 1; i <= n; ++i) fa[i] = i;            int M = (L + R) >> 1;            sort(List + 1, List + M + 1, cmp_a);            bool flag = true;            for (int i = 1; i <= M && flag; ++i) if (i == 1 || List[i].a != List[i - 1].a)            {                for (int j = i; j <= M && flag; ++j) if (j == M || List[j + 1].a != List[i].a)                {                    int left = List[i].x, right = List[i].y;                    for (int k = i + 1; k <= j; ++k)                    {                        left = MAX(left, List[k].x);                        right = MIN(right, List[k].y);                    }                    if (left > right) flag = false;                    if (color(left, right) == false) flag = false;                    for (int k = i; k <= j; ++k) color(List[k].x, List[k].y);                    break;                }            }            if (!flag)                 R = M;            else                L = M + 1;            sort(List + 1, List + M + 1, cmp_id);        }        if (L > m) puts("0"); else printf("%d\n", L);    }    return 0;}