The 37th ACM/ICPC Asia Regional HangZhou Site Online Contest - F
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本题通过率相当低....原因是绝大多数队都是直接贪心吧...找出A最小且B不为0的怪作为入口..可以将所有的B不为0的怪连起来...而中间的间隔可以把 B为0的怪干掉....最后再一一消灭 B 为0的怪...但这种思路是不全面的...如下列数据:
6 20
10 2
10 2
100 0
100 0
100 0
100 0
答案应该是 6 20 先打一个10 2...用2消灭两个100 0...再打一个10 2...用2消灭两个100 0... 但是如果按照裸的贪心..结果会是5 10...从10 2进入..两个10 2 连起来..有3个可用空隙...所以做出来的结果成了5 10
那这个问题如何解决呢? 分为两方面看...若所有B不为0的怪连起来..间隔可以把B为0的都干掉..那么就是最有解了...否则的话就要做一个比较...是按原方案贪心更优..还是说将当前入口点独立出来..其B全都消灭B为0的怪物而不与后面的B不为0的怪相连.再A稍大的做入口再做更优.. .
PS: 比赛的数据太水了..感谢IsaacPei99和luyuncheng提供的数据..掐掉了我比赛时AC的代码..是有问题没考虑全面...代码已更新...
Program:
#include<iostream>#include<stdio.h>#include<algorithm>#include<string.h>#include<math.h>#include<map>#include<queue>#include<stack>#define ll long long#define oo 2000000000#define pi acos(-1) using namespace std; struct node{ ll A,B;}p1[100055];ll n,n1,p0[100055],ss[100055]; bool cmp(node a,node b){ return a.A<b.A;}ll GetMaxNumOfDeleP0(ll m,ll n0){ int l=0,r=n0+1,mid; while (r-l>1) { mid=(r+l)/2; if (ss[mid]>m) r=mid; else l=mid; } return l; }node dfs(node h,ll x,ll m,ll n0){ node q=h; if (m<p1[x].A || x>n1) return h; ll sum=0,i; for (i=x;i<=n1;i++) sum+=p1[i].B-1; sum++; h.A+=n1-x+1; m-=p1[x].A; h.B+=p1[x].A; if (sum>=n0) { h.A+=n0; n0=0; }else { //----------------- q.A+=1+p1[x].B; q.B+=p1[x].A; q=dfs(q,x+1,m,n0-p1[x].B); //----------------- n0-=sum; h.A+=sum; i=GetMaxNumOfDeleP0(m,n0); h.A+=i; h.B+=ss[i]; if (q.A>h.A || q.A==h.A && q.B<h.B) h=q; } return h;}int main(){ ll T,t,i,j,A,B,m,n0; node ans,h; scanf("%I64d",&T); for (t=1;t<=T;t++) { scanf("%I64d%I64d",&n,&m); n1=n0=0; for (i=1;i<=n;i++) { scanf("%I64d%I64d",&A,&B); if (B) { n1++; p1[n1].A=A; p1[n1].B=B; }else p0[++n0]=A; } sort(p1+1,p1+1+n1,cmp); sort(p0+1,p0+1+n0); memset(ss,0,sizeof(ss)); for (i=1;i<=n0;i++) ss[i]=p0[i]+ss[i-1]; ans.A=ans.B=0; ans=dfs(ans,1,m,n0); i=GetMaxNumOfDeleP0(m,n0); h.A=i; h.B=ss[i]; if (h.A>ans.A || h.A==ans.A && h.B<ans.B) ans=h; printf("Case %I64d: %I64d %I64d\n",t,ans.A,ans.B); } return 0;}/*73 54 15 17 72 12 24 0 5 510 14 15 2110 01 03 51 0 2 03 06 2010 210 2100 0100 0100 0100 05 7 7 17 11 01 01 04 54 11 02 01 0*/
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