复习复习DP。。。都忘了QAQ...
好了这道题我其实是看题解才会的。。。
方程 f[i]=min(f[i-j]+v[i]) v[i]表示i是不是石头 s<=j<=t
路径压缩引用一下证明From [email protected]_Hu
#include<cstdio> #include<iostream> #define R register int #include<algorithm> #define R register int const int N=50010,Inf=0x3f3f3f3f; using namespace std; int l,t,s,n,mn=Inf,ans=Inf; int a[N],f[N],d[N],v[N]; inline int g() { R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())); do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret; } inline int min(int a,int b) {return a<b?a:b;} signed main() { l=g(),s=g(),t=g(),n=g(); if(s==t) { R cnt=0; for(R i=1,x;i<=n;++i) x=g(),cnt+=(x%s==0); printf("%d\n",cnt); return 0; } for(R i=1;i<=n;++i) a[i]=g(); sort(a+1,a+n+1); d[n+1]=min(l-a[n],100); l=0; for(R i=1;i<=n;++i) d[i]=min(a[i]-a[i-1],90),l+=d[i],v[l]=1; l+=d[n+1]; for(R i=1;i<=l+10;++i) { f[i]=Inf; for(R j=s;j<=t;++j) if(i>=j) f[i]=min(f[i],f[i-j]+v[i]); } for(R i=l;i<l+10;++i) ans=min(ans,f[i]); printf("%d\n",ans); }
2019.04.28 慌得一批QAQ
原文地址:https://www.cnblogs.com/Jackpei/p/10783092.html
时间: 2024-10-08 20:59:05