hdu 2305 WorstWeather Ever 二分 + 线段树（区间最值）

Problem Description

"Man, this year has the worst weather ever!", David said as he sat crouched in the small cave where we had sought shelter from yet another sudden rainstorm.
"Nuh-uh!", Diana immediately replied in her traditional know-it-all manner.
"Is too!", David countered cunningly. Terrific. Not only were we stuck in this cave, now we would have to listen to those two nagging for at least an hour. It was time to cut this discussion short.
"Big nuh-uh. In fact, 93 years ago it had already rained five times as much by this time of year."
"Duh", David capitulated, "so it's the worst weather in 93 years then."
"Nuh-uh, this is actually the worst weather in 23 years.", Diana again broke in.
"Yeah, well, whatever", David sighed, "Who cares anyway?".
Well, dear contestants, you care, don't you?
Your task is to, given information about the amount of rain during different years in the history of the universe, and a series of statements in the form "Year X had the most rain since year Y", determine whether these are true, might be true, or are false. We say that such a statement is true if:

The amount of rain during these two years and all years between them is known.

It rained at most as much during year X as it did during year Y.

For every year Z satisfying Y < Z < X, the amount of rain during year Z was less than the amount of rain during year X.

We say that such a statement might be true if there is an assignment of amounts of rain to years for which there is no information, such that the statement becomes true. We say that the statement is false otherwise.

 

Input

The input will consist of several test cases, each consisting of two parts.
The first part begins with an integer 1 <= n <= 50000, indicating the number of different years for which there is information. Next follow n lines. The ith of these contains two integers -10⁹ <= yi <= 10⁹ and 1 <= ri <= 10⁹ indicating that there was ri millilitres of rain during year yi (note that the amount of rain during a year can be any nonnegative integer, the limitation on ri is just a limitation on the input). You may assume that yi < yi+1 for 1 <= i < n.
The second part of a test case starts with an integer 1 <= m <= 10000, indicating the number of queries to process. The following m lines each contain two integers -10⁹ <= Y < X <= 10⁹ indicating two years.
There is a blank line between test cases. The input is terminated by a case where n = 0 and m = 0. This case should not be processed.
Technical note: Due to the size of the input, the use of cin/cout in C++ might be too slow in this problem. Use scanf/printf instead. In Java, make sure that both input and output is buffered.

 

Output

There should be m lines of output for each test case, corresponding to the m queries. Queries should be answered with "true" if the statement is true, "maybe" if the statement might be true, and "false" if the statement is false.
Separate the output of two different test cases by a blank line.

 

Sample Input

42002 49202003 59012004 28322005 389022002 20052003 200531985 57821995 30482005 489021985 20052005 201500

Sample Output

falsetruemaybemaybe
题意很啰嗦，大致意思是
给定N(N <= 50000)条信息，表示第yi年的降水量是ri，然后给出M(M <= 10000)
条询问，每条询问的格式是Y X，表示自从第Y年以来X这一年是最大的降水量，问这句
话正确与否。
正确的判断条件是：
1.Y到X这些年的所有年份的降水量已知。
2.Y的降水量 >= X的降水量。
3.对于每个Z，Y < Z < X，Z的降水量小于X的降水量。
可能正确的判断条件是：
其中有一年的降水量不知道。
错误的判断条件是：
其他情况。
#include <iostream>#include<algorithm>using namespace std;#define maxn 50010int n, m;struct point{    int year, r;}pt[maxn];struct Tree{    int Max;    int l, r;}T[maxn*4];int MMax(int a, int b){    return a > b ? a : b;}void Build(int p, int l, int r){    T[p].l = l;    T[p].r = r;    if(l == r)    {        T[p].Max = pt[l].r;        return ;    }    int mid = (l + r) >> 1;    Build(p<<1, l, mid);    Build(p<<1|1, mid+1, r);    T[p].Max = max(T[p<<1].Max, T[p<<1|1].Max);}int Query(int p, int l, int r){    if(r < T[p].l || l > T[p].r)        return 0;    if(l <= T[p].l && T[p].r <= r)        return T[p].Max;    return max(Query(p<<1, l, r), Query(p<<1|1, l, r));}int Binary(int val, int l, int r){    int ans = 0;    while(l <= r) {        int m = (l + r) >> 1;        if(pt[m].year <= val)        {            l = m + 1;            ans = m;        }else            r = m - 1;    }    return ans;}// 连续的块种类int Coces[maxn];int main(){    int i;    int t = 0;    while(scanf("%d", &n) != EOF)    {        if(t++ && n)        {            puts("");        }        for(i = 1; i <= n; i++)        {            scanf("%d %d", &pt[i].year, &pt[i].r);            if(i == 1)            {                Coces[i] = 1;            }            else            {                if(pt[i].year - pt[i-1].year == 1)                    Coces[i] = Coces[i-1];                else                    Coces[i] = Coces[i-1] + 1;            }        }        if(n)            Build(1, 1, n);        scanf("%d", &m);        int bufM = m;        while(bufM--)        {            int Y, X;            int ans; // 0 true 1 maybe 2 false            scanf("%d %d", &Y, &X);            int fY = Binary(Y, 1, n);            int fX = Binary(X, 1, n);            if(pt[fY].year == Y && pt[fX].year == X)            {                // 都能找到数据中有的年份                int Yr = Query(1, fY, fY);                int Zr = Query(1, fY+1, fX-1);                // Y+1 == X 的情况在这里Zr返回的是0，所以肯定满足                int Xr = Query(1, fX, fX);                if(Coces[fY] == Coces[fX])                {                    // 之间的年份全部连续                    if(Yr >= Xr && Zr < Xr)                    {                        ans = 0;                    }                    else                        ans = 2;                }                else                {                    // 之间的年份不连续                    if(Yr >= Xr && Zr < Xr)                    {                        ans = 1;                    }                    else                        ans = 2;                }            }            else if(pt[fX].year == X)            {                // X这一年数据中有                if(Y + 1 == X)                {                    // 当前两年连续                    ans = 1;                }else                {                    int Zr = Query(1, fY+1, fX-1);                    int Xr = Query(1, fX, fX);                    if(Zr < Xr)                        ans = 1;                    else                        ans = 2;                }            }            else if(pt[fY].year == Y)            {                int Yr = Query(1, fY, fY);                int Zr = Query(1, fY+1, fX);                if(Yr > Zr)                {                    ans = 1;                }else                    ans = 2;            }            else            {                // X 和 Y 都没有出现，肯定是maybe                ans = 1;            }            if(!ans)                puts("true");            else if(ans == 1)                puts("maybe");            else                puts("false");        }        if(!n && !m)        {            break;        }    }    return 0;}