# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
504134 | gavgav | Fountain (eJOI20_fountain) | C++17 | 105 ms | 10464 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
// This is Lucallie's solving but with some optimizations and explanations, 'cause when I solved this problem with own code I wondered why this solving is so fast and try to understand it, but he's code is weird and strange, but super fast.
#include <bits/stdc++.h>
using namespace std;
struct reservoir{
int diameter;
short capacity; // 1 <= diameter <= 10^9 | 1 <= capacity <= 1000
};
struct query{
int level, volume, answer; // 1 <= volume <= 10^9
};
int main() {
int height, queries_number, necessary, top, i, j; // 2 <= height <= 10^5 | 1 <= queries 2 * 10^5
scanf( "%d%d", &height, &queries_number);
int previous_request[queries_number + 1], // i-th number of previous_request keeps index of previous request that opened tap of this request
last_request[height + 1] {}, // i-th number of last_request keeps index of last query that open tap at i-th reservoir
necessary_water[height + 1], // i-th number of necessary_water keeps volume of water which need to reach waterways from i-th + 1 reservoir
stream[height + 1];
reservoir reservoirs[height + 1];
query queries[queries_number + 1];
reservoirs[0].diameter = 1000000001; // maximum possible diameter + 1
reservoirs[0].capacity = 0;
for (i = 1; i <= height; ++i)
scanf( "%d%hd", &reservoirs[i].diameter, &reservoirs[i].capacity);
for (i = 1; i <= queries_number; ++i) {
scanf( "%d%d", &queries[i].level, &queries[i].volume);
previous_request[i] = last_request[queries[i].level];
last_request[queries[i].level] = i;
}
stream[0] = 0;
top = 0;
necessary = 0;
int left, middle, right;
for (i = height; i > 0; --i) {
while (reservoirs[i].diameter >= reservoirs[stream[top]].diameter) {
necessary -= reservoirs[stream[top]].capacity;
--top;
}
++top;
stream[top] = i;
necessary_water[top] = necessary;
necessary += reservoirs[i].capacity;
j = last_request[i];
while (j != 0) {
right = top + 1;
left = 0;
while (right - left > 1 ) {
middle = (right + left) / 2;
if (necessary - necessary_water[middle] >= queries[j].volume)
left = middle;
else
right = middle;
}
queries[j].answer = stream[left];
j = previous_request[j];
}
}
for (i = 1; i <= queries_number; ++i){
printf("%d\n", queries[i].answer);
}
}
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