| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 2616 | Accepted: 925 |
Description
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
一道有点难的贪心加优先队列问题
具体思路在代码的注释中有详细说明
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 5 using namespace std; 6 7 typedef struct 8 { 9 int minspf; 10 int maxspf; 11 } COW; 12 13 typedef struct 14 { 15 int spf; 16 int cover; 17 } LOTION; 18 19 int c,l,sz=0; 20 COW cow[2550],heap[2550]; 21 LOTION lotion[2550]; 22 23 bool cmp_cow(COW a,COW b) 24 { 25 return a.minspf<b.minspf; 26 } 27 28 bool cmp_lotion(LOTION a,LOTION b) 29 { 30 return a.spf<b.spf; 31 } 32 33 //对牛维护一个小根堆，键值为所需防晒霜最大能力 34 void push(COW x) 35 { 36 int i=sz++; 37 while(i>0) 38 { 39 int p=(i-1)/2; 40 if(heap[p].maxspf<=x.maxspf) 41 break; 42 heap[i]=heap[p]; 43 i=p; 44 } 45 heap[i]=x; 46 } 47 48 COW pop() 49 { 50 COW ret=heap[0]; 51 COW x=heap[--sz]; 52 int i=0; 53 while(i*2+1<=sz) 54 { 55 int a=i*2+1,b=i*2+2; 56 if(b<sz&&heap[b].maxspf<heap[a].maxspf) 57 a=b; 58 if(heap[a].maxspf>=x.maxspf) 59 break; 60 heap[i]=heap[a]; 61 i=a; 62 } 63 heap[i]=x; 64 return ret; 65 } 66 67 int main() 68 { 69 while(scanf("%d %d",&c,&l)==2) 70 { 71 sz=0; 72 for(int i=0;i<c;i++) 73 scanf("%d %d",&cow[i].minspf,&cow[i].maxspf); 74 for(int i=0;i<l;i++) 75 scanf("%d %d",&lotion[i].spf,&lotion[i].cover); 76 sort(cow,cow+c,cmp_cow);//将所有牛按它们所需防晒霜的最小防晒能力排序 77 sort(lotion,lotion+l,cmp_lotion);//将将所有防晒霜按它们的防晒能力排序 78 int ans=0,t=0; 79 for(int i=0;i<l;i++) 80 { 81 while(t<c&&lotion[i].spf>=cow[t].minspf)//对于每一种防晒霜，先将所有最小值比它大的牛插入堆中 82 { 83 push(cow[t]); 84 t++; 85 } 86 while(sz>0&&lotion[i].spf>heap[0].maxspf)//再将所有最大值比它小的牛出堆 87 pop(); 88 while(lotion[i].cover>0&&sz>0)//最后将所有它能保护的牛出堆并将答案+1 89 { 90 pop(); 91 ans++; 92 lotion[i].cover--; 93 } 94 } 95 printf("%d ",ans); 96 } 97 98 return 0; 99 }