SMU 2023 Spring 题单 第二周 贪心
Saruman's Army
首先对序列排序，然后逐个考虑覆盖，如果要覆盖当前的点，则标记点越靠后越好，所有向后找\(R\)，选择最靠后的标记，然后从标记点开始在向后找\(R\)也是被标记过的，直接跳过
#include <cstdio>
#include <algorithm>
using namespace std;
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
const int N = 1005;
int n, r, a[N], cnt;
int main() {
while (true) {
r = read(), n = read(), cnt = 0;
if (r == n && r == -1) return 0;
for (int i = 1; i <= n; i++) a[i] = read();
sort( a + 1 , a + 1 + n );
for( int i = 1 , j , k ; i <= n ; i = k ){
cnt ++ ;
for( j = i ; j <= n && a[j] <= a[i] + r ; j ++ );
j --;
for( k = j ; k <= n && a[k] <= a[j] + r ; k ++ );
}
printf("%d\n" , cnt );
}
return 0;
}
Fence Repair
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
#define int long long
signed main() {
int n , res = 0;
cin >> n;
priority_queue<int,vector<int> , greater<int> > q;
for( int i = 1 , x ; i <= n ; i ++ )
cin >> x , q.push(x);
for( int a , b ; q.size() > 1; ){
a = q.top() , q.pop();
b = q.top() , q.pop();
res += a + b , q.push( a + b );
}
cout << res << "\n";
return 0;
}
Protecting the Flowers
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
#define int long long
struct node{
int t , d;
};
bool cmp( node a , node b ){
return a.t * b.d < b.t * a.d;
}
signed main() {
int n , res = 0 , cnt = 0;
cin >> n;
vector<node> a(n);
for( int i = 0 ; i < n ; i ++ )
cin >> a[i].t >> a[i].d;
sort( a.begin() , a.end() , cmp );
for( int i = 0 ; i < n ; i ++ ){
res += cnt * a[i].d , cnt += 2ll * a[i].t;
}
cout << res;
return 0;
}
Yogurt factory
第i天生产酸奶的成本是c[i] = min( c[i] , c[i-1] + s )
求出每天最低酸奶成本后直接计算就好了。
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
#define int long long
signed main() {
int n , s , res = 0;
cin >> n >> s;
vector<int> c(n+1) , y( n+1 );
for( int i = 1 ; i <= n ; i ++ )
cin >> c[i] >> y[i];
for( int i = 2 ; i <= n ; i ++ )
c[i] = min( c[i] , c[i-1] + s );
for( int i = 1 ; i <= n ; i ++ )
res += c[i] * y[i];
cout << res;
return 0;
}
Cleaning Shifts
其实首先我们把所有的区间进行排序，首先安装左端点排序从小到大排序，如果左端点相同按照右端点从大到小排序。
排序后，我们在保证做端点可以衔接上的情况下尽可能大选择右端点即可。
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct node {
int l, r;
};
bool cmp(node a, node b) {
if (a.l != b.l) return a.l < b.l;
else return a.r > b.r;
}
int main() {
int n, t;
while (cin >> n >> t) {
vector<node> a(n);
for (int i = 0; i < n; i++)
cin >> a[i].l >> a[i].r;
sort(a.begin(), a.end(), cmp);
if (a[0].l > 1) {
cout << "-1\n";
continue;
}
int res = 1, last = a[0].r, pos = last;
for (int i = 1; i < n; i++) {
if (a[i].l <= last + 1) { // 可以衔接上
pos = max( pos , a[i].r);
if (i == n - 1) {
if (pos > last) last = pos, res++;
break;
}
} else { // 衔接不上了
if (pos > last) { // 有新增的覆盖
last = pos, res++;
if (last == t) break;
i--; // 当前区间还没有判断过
} else {
res = -1;
break;// 无新增覆盖，说明之后的都衔接不上了
}
}
}
if (last != t) cout << "-1\n";
else cout << res << "\n";
}
return 0;
}
Packets
优先放3,4,5,6然后用 2 去插空，剩余的 2 再占用新的盒子，最后用 1 去插空，剩余的 1 再占用新的盒子
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <cstdio>
using namespace std;
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
typedef pair<int, int> pii;
typedef pair<int, pii> piii;
int main() {
int a, b, c, d, e, f;
int A, B, cnt;
while (true) {
a = read(), b = read(), c = read(), d = read(), e = read(), f = read();
if (a == 0 && b == 0 && c == 0 && d == 0 && e == 0 && f == 0)
return 0;
A = B = cnt = 0;
cnt = (c + 3) / 4 + d + e + f;
B = d * 5;
if (c % 4 == 1) B += 5;
else if (c % 4 == 2) B += 3;
else if (c % 4 == 3) B += 1;
cnt += (max(0, b - B) + 8) / 9;
A = cnt * 36 - b * 4 - c * 9 - d * 16 - e * 25 - f * 36;
cnt += (max(0, a - A) + 35) / 36;
cout << cnt << "\n";
}
return 0;
}
Allowance
我们把所有的硬币按照面值排序，然后从大到小选择，在保证当前方案不超过C的情况下尽可能的多选面值大的。然后在从小到大选择，在保证可以大于等于C的情况下尽可能的选择面值小的。然后计算出当前方案可以使用的天数，如果无法构成，说明已经无法发放工资。
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <cstdio>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pii;
int main() {
ll n, c, res = 0;
cin >> n >> c;
vector<pii> coin;
for (int i = 1, v, b; i <= n; i++) {
cin >> v >> b;
if (v >= c) res += b;
else coin.push_back({v, b});
}
sort(coin.begin(), coin.end());
while (true) {
vector<int> used(coin.size(), 0);
int cnt = c;
for (int i = coin.size() - 1, t; i >= 0 && cnt > 0; i--) {
t = min(cnt / coin[i].first, coin[i].second);
used[i] = t, cnt -= t * coin[i].first;
}
for (int i = 0; cnt > 0 && i < coin.size(); i++)
while (cnt > 0 && used[i] < coin[i].second)
used[i]++, cnt -= coin[i].first;
if( cnt > 0 ) break;
ll ans = 1e18;
for( int i = 0 ; i < coin.size() ; i ++ )
if( used[i] > 0 ) ans = min( ans , coin[i].second / used[i] );
res += ans;
for( int i = 0 ; i < coin.size() ; i ++ )
coin[i].second -= ans * used[i];
}
cout << res << "\n";
return 0;
}
Stripies
这种一眼贪心，其实最简单的方法就是两种贪心策略都试就好。
证明其实也不难，要把\(a,b,c\)合成\(d\)
\[d = 2\sqrt{2\sqrt{ab}c}\\
\frac{d^2}{8}=\sqrt {a} \sqrt {b} c
\]
显然这里，\(d>0\)恒成立，所以\(d\)和\(\frac{d^2}{8}\)增减性相同，要使得\(d\)最小，则\(a,b\)是较大的两个值
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long ll;
int main() {
priority_queue<double> q;
int n;
cin >> n;
for (double x; n; n--)
cin >> x, q.push(x);
for (double a, b, c; q.size() > 1;) {
a = q.top(), q.pop();
b = q.top(), q.pop();
c = 2.0 * sqrt(a * b) , q.push(c);
}
printf("%.3f" , q.top() );
return 0;
}
