CF1498-B.Box Fitting
作者:互联网
CF1498-B.Box Fitting
题意
给出 n n n个长条,每个长条保证可以表示为 2 x 2^x 2x的形式,问你如果一个宽度为 w w w的盒子最少要多高才能装下这些长条。
思路
贪心。将长条按照长度从大到小排序,对于每一层我们尽量将它装满再装下一层。
可以用 m u l t i s e t multiset multiset维护每一层剩余的空间。对于当前要放入盒子的长条,在集合中 l o w e r _ b o u n d lower\_bound lower_bound得到一个合适的位置,如果没有那就新开一层,答案加一;如果有就将该层的空间减少,还有剩余就再次加入到集合中。
代码
#include <algorithm>
#include <iostream>
#include <cstring>
#include <stack>
#include <cstdlib>
#include <map>
#include <queue>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <set>
#include <ctime>
#define rep(i, s, t) for (int i = (s); i < (t); i++)
#define wa std::cerr << "----WARN----" << std::endl;
#define wl std::cerr << "********" << std::endl;
#define wr std::cerr << "~~~~~~~~" << std::endl;
#define PII pair<int, int>
#define mp(a, b) make_pair(a, b)
#define fr first
#define sc second
typedef long long LL;
const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const LL LINF = 1LL * 0x3f3f3f3f3f3f3f3f;
inline int read();
const int N = 100005;
int a[N];
void solve() {
int n, w;
std::cin >> n >> w;
rep (i, 0, n) {
std::cin >> a[i];
}
std::sort(a, a + n, std::greater<int>());
std::multiset<int> s;
int ans = 0;
rep (i, 0, n) {
if (s.empty()) {
if (a[i] != w) s.insert(w - a[i]);
ans++;
} else {
auto it = s.lower_bound(a[i]);
if (it == s.end()) {
if (w != a[i]) s.insert(w - a[i]);
ans++;
continue;
} else {
int t = *it;
s.erase(it);
if (t != a[i]) s.insert(t - a[i]);
}
}
}
std::cout << ans << std::endl;
}
inline int read() {
int s = 0, w = 1;
char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') w = -1;
ch = getchar();
}
while (isdigit(ch)) {
s = s * 10 + ch - '0';
ch = getchar();
}
return s * w;
}
signed main() {
// freopen("in.txt", "r", stdin);
int T = 1;
T = read();
while (T--) solve();
return 0;
}
标签:Box,std,ch,int,Fitting,CF1498,include,长条,define 来源: https://blog.csdn.net/m0_47660709/article/details/117186761