不得不說是國文題,題目很短,條件略多。
貌似網上大部分人都是些RMQ的,我寫了個單調棧。不過速度上并沒有RMQ快(跑了300+ms),不知道是不是代碼寫渣了。
大概的思路就是維護每一年的 第一個大于等于當年的降水量的年份(往年 和 未來 的年份都要,是以要做兩遍),時間複雜度O(n)。
具體的分析可見huzecong的分析,寫的很詳細。
需要注意的是年份的範圍是-10^9<=yi<=10^9,棧的初始化要是-INF。
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
const int MAXN = 50010;
const int INF = 1999999999;
int a[MAXN], b[MAXN], c[MAXN], year[MAXN], rain[MAXN], n;
struct stack
{
int y, v;
}s[MAXN],cur;
bool find(int x, int &xx)
{
int l = 1, r = n, mid;
while (l <= r)
{
mid = (l + r) >> 1;
if (year[mid] == x)
{
xx = mid;
return 1;
}
else if (year[mid] > x) r = mid - 1;
else l = mid + 1;
}
while (year[r] >= x) --r;
xx = r;
return 0;
}
void work()
{
int i, j, m, top = 0, last = 0, x, y, xx, yy;
bool b1, b2;
scanf("%d", &n);
b[0] = s[0].y = s[0].v = -INF;
for (i = 1; i <= n; ++i)
{
scanf("%d%d", &cur.y, &cur.v);
year[i] = cur.y; rain[i] = cur.v;
while (top&&s[top].v < cur.v) --top;
a[i] = s[top].y;
s[++top] = cur;
if (year[i - 1] + 1 != year[i]) b[i] = year[i] - 1;
else b[i] = b[i - 1];
}
top = 0; s[0].y = INF;
for (i = n; i; --i)
{
cur.y = year[i]; cur.v = rain[i];
while (top&&s[top].v < cur.v) --top;
c[i] = s[top].y;
s[++top] = cur;
}
scanf("%d", &m);
for (i = 1; i <= m; ++i)
{
scanf("%d%d", &y, &x);
b1 = find(x, xx);
b2 = find(y, yy);
if (b1&&b2)
{
if (a[xx] != y) printf("false\n");
else if (b[xx] >= y) printf("maybe\n");
else printf("true\n");
}
if (b1&&!b2)
{
if (a[xx] > y) printf("false\n");
else printf("maybe\n");
}
if (!b1&&b2)
{
if (year[xx] == y || (c[yy] >= x&&rain[xx] < rain[yy])) printf("maybe\n");
else printf("false\n");
}
if (!b1&&!b2) printf("maybe\n");
}
}
int main()
{
work();
return 0;
}