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GD5: Scheduling on many machines
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#include <bits/stdc++.h> | |
using namespace std; | |
void solve() | |
{ | |
int n, m; | |
scanf("%d %d", &n, &m); | |
priority_queue<int, vector<int>, greater<int>> pq; | |
for (int i = 0; i < m; i++) // init m machines | |
pq.push(0); | |
for (int i = 0; i < n; i++) { // process jobs | |
int inp; | |
scanf("%d", &inp); | |
int cur = pq.top(); | |
pq.pop(); | |
pq.push(cur + inp); | |
} | |
int ans = 0; | |
while (pq.size() > 0) { | |
ans = max(ans, pq.top()); | |
pq.pop(); | |
} | |
printf("%d\n", ans); | |
} | |
int main() | |
{ | |
int ncase; | |
scanf("%d", &ncase); | |
while (ncase--) | |
solve(); | |
return 0; | |
} |
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#include <bits/stdc++.h> | |
using namespace std; | |
// min heap | |
struct my_pq { | |
int nxt = 0; | |
vector<int> data; | |
void clear() | |
{ | |
data.clear(); | |
nxt = 0; | |
} | |
int top() | |
{ | |
return data[0]; | |
} | |
void pop() | |
{ | |
if (nxt == 1) | |
nxt = 0; | |
else { | |
data[0] = data[nxt - 1]; | |
nxt--; | |
int cur = 0; | |
while (cur < nxt) { | |
/* | |
a | |
/ \ | |
b c | |
a < b and a < c relation! Tricky to implement! | |
*/ | |
int change = cur; | |
int l = lc(cur); | |
if (l < nxt && data[l] < data[change]) | |
change = l; | |
int r = rc(cur); | |
if (r < nxt && data[r] < data[change]) | |
change = r; | |
if (change != cur) { | |
swap(data[cur], data[change]); | |
if (change == l) | |
cur = l; | |
else | |
cur = r; | |
} else | |
break; | |
} | |
} | |
} | |
int par(int x) | |
{ | |
if (x % 2 == 0) | |
return (x - 2) / 2; | |
return (x - 1) / 2; | |
} | |
int lc(int x) | |
{ | |
return x * 2 + 1; | |
} | |
int rc(int x) | |
{ | |
return x * 2 + 2; | |
} | |
void push(int x) | |
{ | |
if (nxt == (int)data.size()) | |
data.push_back(x); | |
else | |
data[nxt] = x; | |
nxt++; | |
int idx = (int)data.size() - 1; | |
while (idx > 0) { | |
int p = par(idx); | |
if (data[p] > data[idx]) { | |
swap(data[p], data[idx]); | |
idx = p; | |
} else | |
break; | |
} | |
} | |
int size() | |
{ | |
return nxt; | |
} | |
} pq; | |
void solve() | |
{ | |
int n, m; | |
scanf("%d %d", &n, &m); | |
pq.clear(); | |
for (int i = 0; i < m; i++) // init m machines | |
pq.push(0); | |
for (int i = 0; i < n; i++) { // process jobs | |
int inp; | |
scanf("%d", &inp); | |
int cur = pq.top(); | |
pq.pop(); | |
pq.push(cur + inp); | |
} | |
int ans = 0; | |
while (pq.size() > 0) { | |
ans = max(ans, pq.top()); | |
pq.pop(); | |
} | |
printf("%d\n", ans); | |
} | |
int main() | |
{ | |
int ncase; | |
scanf("%d", &ncase); | |
while (ncase--) | |
solve(); | |
return 0; | |
} |
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