GYM 101149 G.Of Zorcs and Axes(set)
2017-03-19 10:14
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Description
从m个物品中选出n个,m个物品每个有重量w[i],价值c[i],要求选出的第i个物品的重量不小于a[i],价值不小于b[i],问是否有合法方案
Input
第一行一整数n表示要选出物品数量,之后n行每行两个整数a[i]和b[i]表示对选出的第i个物品的限制,之后输入一整数m表示所有物品数,之后m行每行两个整数w[i]和c[i]表示第i件物品的重量和价值(1<=n,m<=2e5,1<=a[i],b[i],w[i],c[i]<=1e9)
Output
如果存在合法方案则输出n个整数表示选出的第i个物品的编号,否则输出-1
Sample Input
3
3 4
5 1
2 6
4
5 7
4 5
3 3
5 2
Sample Output
2 4 1
Solution
对n个限制按b[i]降序排,对m件物品按c[i]降序排,游标法,每次把c[i]大于等于当前限制b[i]的物品拿出来,找一个重量大于等于a[i]的最小值作为这个限制选出的物品,用set维护当前可以用的物品,每次二分查找重量大于等于a[i]的物品,时间复杂度O(mlogm+nlogn)
Code
从m个物品中选出n个,m个物品每个有重量w[i],价值c[i],要求选出的第i个物品的重量不小于a[i],价值不小于b[i],问是否有合法方案
Input
第一行一整数n表示要选出物品数量,之后n行每行两个整数a[i]和b[i]表示对选出的第i个物品的限制,之后输入一整数m表示所有物品数,之后m行每行两个整数w[i]和c[i]表示第i件物品的重量和价值(1<=n,m<=2e5,1<=a[i],b[i],w[i],c[i]<=1e9)
Output
如果存在合法方案则输出n个整数表示选出的第i个物品的编号,否则输出-1
Sample Input
3
3 4
5 1
2 6
4
5 7
4 5
3 3
5 2
Sample Output
2 4 1
Solution
对n个限制按b[i]降序排,对m件物品按c[i]降序排,游标法,每次把c[i]大于等于当前限制b[i]的物品拿出来,找一个重量大于等于a[i]的最小值作为这个限制选出的物品,用set维护当前可以用的物品,每次二分查找重量大于等于a[i]的物品,时间复杂度O(mlogm+nlogn)
Code
#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<vector> #include<queue> #include<map> #include<set> #include<ctime> using namespace std; typedef long long ll; #define INF 0x3f3f3f3f #define maxn 222222 int n,m,ans[maxn]; struct node { int w,c,id; bool operator<(const node&b)const { return w>b.w; } }a[maxn],b[maxn]; typedef pair<int,int>P; set<P>s; set<P>::iterator it; int main() { while(~scanf("%d",&n)) { for(int i=1;i<=n;i++)scanf("%d%d",&a[i].w,&a[i].c),a[i].id=i; scanf("%d",&m); for(int i=1;i<=m;i++)scanf("%d%d",&b[i].w,&b[i].c),b[i].id=i; sort(a+1,a+n+1); sort(b+1,b+m+1); s.clear(); int i=1,j=1,gg=0; while(i<=n) { while(j<=m&&b[j].w>=a[i].w)s.insert(P(b[j].c,b[j].id)),j++; it=s.lower_bound(P(a[i].c,-1)); if(it==s.end()) { gg=1; break; } ans[a[i].id]=(*it).second; s.erase(it); i++; } if(gg)printf("-1\n"); else { for(int i=1;i<=n;i++)printf("%d%c",ans[i],i==n?'\n':' '); } } return 0; }
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