luogu2678 [NOIp2015]跳石头 (二分答案+贪心)

先二分出一个x，我们要算使最近的跳跃距离>=x的最少移除数量是否<=M就可以了

然后就别dp了...贪心就完事了...我肯定能不移就不移比较好...

#include<bits/stdc++.h>
#define pa pair<int,int>
#define lowb(x) ((x)&(-(x)))
#define REP(i,n0,n) for(i=n0;i<=n;i++)
#define PER(i,n0,n) for(i=n;i>=n0;i--)
#define MAX(a,b) ((a>b)?a:b)
#define MIN(a,b) ((a<b)?a:b)
#define CLR(a,x) memset(a,x,sizeof(a))
#define rei register int
using namespace std;
typedef long long ll;
const int maxn=50050,inf=1e9+1;

inline ll rd(){
    ll x=0;char c=getchar();int neg=1;
    while(c<'0'||c>'9'){if(c=='-') neg=-1;c=getchar();}
    while(c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
    return x*neg;
}

int L,N,M,dis[maxn];

inline bool judge(int m){
    int x=0,y=0,i=0;
    while(y<N){
        x=y;
        for(y++;dis[y]-dis[x]<m&&y<N;y++,i++);
    }
    return L-dis[x]>=m&&i<=M;
}

int main(){
    //freopen(".in","r",stdin);
    rei i,j,k;
    L=rd(),N=rd(),M=rd();
    for(i=1;i<=N;i++) dis[i]=rd();
    dis[++N]=L;
    int l=1,r=L,ans;
    while(l<=r){
        int m=l+r>>1;
        if(judge(m)) ans=m,l=m+1;
        else r=m-1;
    }
    printf("%d
",ans);
    return 0;
}