POJ - 3614的贪心算法如何改写成长尾词？
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本文共计738个文字，预计阅读时间需要3分钟。
为避免晒伤，每头C（1≤C≤2500）只奶牛在海滩上时必须用防晒霜涂抹其皮肤。奶牛i有最低和最高防晒指数（SPF）限制（1≤minSPFi≤1000；minSPFi≤maxSPFi≤1000）。
To avoid unsightly burns while tanning, each of the C (1 ≤ C ≤ 2500) cows must cover her hide with sunscreen when they’re at the beach. Cow i has a minimum and maximum SPF rating (1 ≤ minSPFi ≤ 1,000; minSPFi ≤ maxSPFi ≤ 1,000) that will work. If the SPF rating is too low, the cow suffers sunburn; if the SPF rating is too high, the cow doesn’t tan at all…
The cows have a picnic basket with L (1 ≤ L ≤ 2500) bottles of sunscreen lotion, each bottle i with an SPF rating SPFi (1 ≤ SPFi ≤ 1,000). Lotion bottle i can cover coveri cows with lotion. A cow may lotion from only one bottle.
What is the maximum number of cows that can protect themselves while tanning given the available lotions?
Input
- Line 1: Two space-separated integers: C and L
- Lines 2…C+1: Line i describes cow i’s lotion requires with two integers: minSPFi and maxSPFi
- Lines C+2…C+L+1: Line i+C+1 describes a sunscreen lotion bottle i with space-separated integers: SPFi and coveri
Output
A single line with an integer that is the maximum number of cows that can be protected while tanning
Sample Input
3 2
3 10
2 5
1 5
6 2
4 1
Sample Output
2
思路：将每个奶牛的minf按降序排序，然后在合法的情况下选择最大spf的防晒霜，
由贪心可知正确性显然
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<sstream>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 600005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-6
#define pll pair<ll,ll>
ll quickpow(ll a, ll b) {
ll ans = 1;
a = a % mod;
while (b > 0) {
if (b % 2)ans = ans * a;
b = b / 2;
a = a * a;
}
return ans;
}
int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a%b);
}
struct Cow {
int minf, maxf;
}cow[maxn];
bool cmp(Cow a, Cow b) {
return a.minf > b.minf;
}
int spf[maxn], cover[maxn];
int main()
{
ios::sync_with_stdio(false);
int c, l;
cin >> c >> l;
int i, j;
for (i = 1; i <= c; i++) {
cin >> cow[i].minf >> cow[i].maxf;
}
sort(cow + 1, cow + 1 + c, cmp);
int tot = 0;
for (i = 1; i <= l; i++) {
cin >> spf[i] >> cover[i];
}
int pos;
for (i = 1; i <= c; i++) {
int maxx = -1;
pos = -1;
for (j = 1; j <= l; j++) {
if (maxx < spf[j]) {
if (spf[j] >= cow[i].minf&&spf[j] <= cow[i].maxf&&cover[j] >= 1) {
maxx = spf[j];
pos = j;
}
}
}
if (pos != -1) {
cover[pos]--;
tot++;
}
}
cout << tot << endl;
}
本文共计738个文字，预计阅读时间需要3分钟。
为避免晒伤，每头C（1≤C≤2500）只奶牛在海滩上时必须用防晒霜涂抹其皮肤。奶牛i有最低和最高防晒指数（SPF）限制（1≤minSPFi≤1000；minSPFi≤maxSPFi≤1000）。
To avoid unsightly burns while tanning, each of the C (1 ≤ C ≤ 2500) cows must cover her hide with sunscreen when they’re at the beach. Cow i has a minimum and maximum SPF rating (1 ≤ minSPFi ≤ 1,000; minSPFi ≤ maxSPFi ≤ 1,000) that will work. If the SPF rating is too low, the cow suffers sunburn; if the SPF rating is too high, the cow doesn’t tan at all…
The cows have a picnic basket with L (1 ≤ L ≤ 2500) bottles of sunscreen lotion, each bottle i with an SPF rating SPFi (1 ≤ SPFi ≤ 1,000). Lotion bottle i can cover coveri cows with lotion. A cow may lotion from only one bottle.
What is the maximum number of cows that can protect themselves while tanning given the available lotions?
Input
- Line 1: Two space-separated integers: C and L
- Lines 2…C+1: Line i describes cow i’s lotion requires with two integers: minSPFi and maxSPFi
- Lines C+2…C+L+1: Line i+C+1 describes a sunscreen lotion bottle i with space-separated integers: SPFi and coveri
Output
A single line with an integer that is the maximum number of cows that can be protected while tanning
Sample Input
3 2
3 10
2 5
1 5
6 2
4 1
Sample Output
2
思路：将每个奶牛的minf按降序排序，然后在合法的情况下选择最大spf的防晒霜，
由贪心可知正确性显然
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<sstream>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 600005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-6
#define pll pair<ll,ll>
ll quickpow(ll a, ll b) {
ll ans = 1;
a = a % mod;
while (b > 0) {
if (b % 2)ans = ans * a;
b = b / 2;
a = a * a;
}
return ans;
}
int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a%b);
}
struct Cow {
int minf, maxf;
}cow[maxn];
bool cmp(Cow a, Cow b) {
return a.minf > b.minf;
}
int spf[maxn], cover[maxn];
int main()
{
ios::sync_with_stdio(false);
int c, l;
cin >> c >> l;
int i, j;
for (i = 1; i <= c; i++) {
cin >> cow[i].minf >> cow[i].maxf;
}
sort(cow + 1, cow + 1 + c, cmp);
int tot = 0;
for (i = 1; i <= l; i++) {
cin >> spf[i] >> cover[i];
}
int pos;
for (i = 1; i <= c; i++) {
int maxx = -1;
pos = -1;
for (j = 1; j <= l; j++) {
if (maxx < spf[j]) {
if (spf[j] >= cow[i].minf&&spf[j] <= cow[i].maxf&&cover[j] >= 1) {
maxx = spf[j];
pos = j;
}
}
}
if (pos != -1) {
cover[pos]--;
tot++;
}
}
cout << tot << endl;
}