1067: [SCOI2007]降雨量
Time Limit: 1 Sec Memory Limit: 162 MBSubmit: 5674 Solved: 1484
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Description
我們常常會說這樣的話:“X年是自Y年以來降雨量最多的”。它的含義是X年的降雨量不超過Y年,且對於任意
Y<Z<X,Z年的降雨量嚴格小於X年。例如2002,2003,2004和2005年的降雨量分別爲4920,5901,2832和3890,
則可以說“2005年是自2003年以來最多的”,但不能說“2005年是自2002年以來最多的”由於有些年份的降雨量未
知,有的說法是可能正確也可以不正確的。
Input
輸入僅一行包含一個正整數n,爲已知的數據。以下n行每行兩個整數yi和ri,爲年份和降雨量,按照年份從小
到大排列,即yi<yi+1。下一行包含一個正整數m,爲詢問的次數。以下m行每行包含兩個數Y和X,即詢問“X年是
自Y年以來降雨量最多的。”這句話是必真、必假還是“有可能”。
Output
對於每一個詢問,輸出true,false或者maybe。
Sample Input
2002 4920
2003 5901
2004 2832
2005 3890
2007 5609
2008 3024
5
2002 2005
2003 2005
2002 2007
2003 2007
2005 2008
Sample Output
true
false
maybe
false
HINT
100%的數據滿足:1<=n<=50000, 1<=m<=10000, -10^9<=yi<=10^9, 1<=ri<=10^9
#include <bits/stdc++.h>
#define pb push_back
using namespace std;
const int N = 100010;
const int INF = 1e9 + 7;
struct Tree
{
int l;
int r;
int maxx;
int minn;
};
Tree tree[N * 4];
int n, m;
int ma[N];
int mb[N];
int cnt[N];
vector<int> id;
void up(int x)
{
tree[x].maxx = max(tree[x << 1].maxx, tree[x << 1 | 1].maxx);
tree[x].minn = min(tree[x << 1].minn, tree[x << 1 | 1].minn);
}
void build(int x, int l, int r)
{
//printf("x = %d l = %d r = %d\n", x, l, r);
tree[x].l = l;
tree[x].r = r;
tree[x].maxx = -1;
tree[x].minn = INF;
if(l == r){
tree[x].maxx = cnt[l];
tree[x].minn = cnt[l];
}
else{
int mid;
mid = l + ((r - l) >> 1);
build(x << 1, l, mid);
build(x << 1 | 1, mid + 1, r);
up(x);
}
}
int quary1(int x, int l, int r)
{
//printf("%d\n", x);
int ll = tree[x].l;
int rr = tree[x].r;
if(l > r){
return -1;
}
if(ll == l && rr == r){
return tree[x].maxx;
}
else{
int ans = -1;
int mid;
mid = ll + ((rr - ll) >> 1);
ans = max(ans, quary1(x << 1, l, min(r, mid)));
ans = max(ans, quary1(x << 1 | 1, max(mid + 1, l), r));
return ans;
}
}
int quary2(int x, int l, int r)
{
//printf("%d\n", x);
int ll = tree[x].l;
int rr = tree[x].r;
if(l > r){
return INF;
}
if(ll == l && rr == r){
//printf("l = %d r = %d min = %d\n", l, r, tree[x].minn);
return tree[x].minn;
}
else{
int ans = INF;
int mid;
mid = ll + ((rr - ll) >> 1);
ans = min(ans, quary2(x << 1, l, min(r, mid)));
// if(ans == -1){
// printf("l = %d min(r, mid) = %d\n", l, min(r, mid));
// //getchar();
// }
ans = min(ans, quary2(x << 1 | 1, max(mid + 1, l), r));
// if(ans == -1){
// printf("max(mid + 1, l) = %d r = %d\n", max(mid + 1, l), r);
// //getchar();
// }
return ans;
}
}
int get_id(int x)
{
return lower_bound(id.begin(), id.end(), x) - id.begin() + 1;
}
int main(int argc, char const *argv[])
{
int len;
scanf("%d", &n);
id.clear();
memset(cnt, -1, sizeof(cnt));
for(int i = 0; i < n; i ++){
scanf("%d%d", &ma[i], &mb[i]);
id.pb(ma[i]);
}
sort(id.begin(), id.end());
id.erase(unique(id.begin(), id.end()), id.end());
len = id.size();
for(int i = 1; i < len; i ++){
if(id[i] != id[i - 1] + 1){
id.pb(id[i - 1] + 1);
}
}
sort(id.begin(), id.end());
len = id.size();
//printf("len = %d\n", len);
// for(int i = 0; i < len; i ++){
// cout << id[i] << endl;
// }
for(int i = 0; i < n; i ++){
cnt[get_id(ma[i])] = mb[i];
}
// for(int i = 1; i <= len; i ++){
// printf("%d - %d - %d\n", i, id[i - 1], cnt[i]);
// }
// printf("\n");
build(1, 1, len);
// for(int i = 1; i <= 15; i ++){
// printf("x = %d l = %d r = %d maxx = %d minn = %d\n", i, tree[i].l, tree[i].r, tree[i].maxx, tree[i].minn);
// }
scanf("%d", &m);
for(int i = 0; i < m; i ++){
int ll, rr;
int l, r;
int be;
int en;
int t, q;
scanf("%d%d", &ll, &rr);
l = get_id(ll);
r = get_id(rr);
be = cnt[l];
en = cnt[r];
if(ll != id[l - 1]){
l --;
be = -1;
}
if(id[r - 1] != rr){
en = -1;
}
t = quary1(1, l + 1, r - 1);
q = quary2(1, l + 1, r - 1);
if(be >= en && t < en && q != -1 && be != -1 && en != -1){
printf("true\n");
}
else if((be < en && be != -1 && en != -1 ) || (t >= be && be != -1) || (t >= en && en != -1)){
printf("false\n");
}
else{
printf("maybe\n");
}
}
return 0;
}
/*
6
7880 5614
4887 4971
8113 2965
320 5697
6066 6187
5058 3536
1
5422 7880
*/