还是水题一个,对两个有牛的牛棚之间进行贪心,然后对长度为S的木板递减,直到M用光或者没有间距;
code
/* ID: yueqiq PROG: barn1 LANG: C++ */ #include <set> #include <map> #include <ctime> #include <queue> #include <cmath> #include <stack> #include <limits> #include <vector> #include <bitset> #include <string> #include <cstdio> #include <cstring> #include <fstream> #include <string.h> #include <iostream> #include <algorithm> #define Si set<int> #define LL long long #define pb push_back #define PS printf(" ") #define Vi vector<int> #define LN printf("\n") #define lson l,m,rt << 1 #define rson m+1,r,rt<<1|1 #define SD(a) scanf("%d",&a) #define PD(a) printf("%d",a) #define SET(a,b) memset(a,b,sizeof(a)) #define FF(i,a) for(int i(0);i<(a);i++) #define FD(i,a) for(int i(a);i>=(1);i--) #define FOR(i,a,b) for(int i(a);i<=(b);i++) #define FOD(i,a,b) for(int i(a);i>=(b);i--) #define readf freopen("barn1.in","r",stdin) #define writef freopen("barn1.out","w",stdout) const int maxn = 201; const int INF = 0x3fffffff; const int dx[]={0,1,0,-1}; const int dy[]={1,0,-1,0}; const double pi = acos(-1.0); using namespace std; int M,S,C; bool use[maxn]; int dis[maxn]; int main(){ readf; writef; scanf("%d%d%d",&M,&S,&C); int mm=INF,mx=0; FOR(i,1,C){ int k;SD(k); if(k<mm) mm=k; if(k>mx) mx=k; use[k]=true; } int tmp=mm; int cnt=0; FOR(i,mm+1,mx){ if(use[i]){ dis[cnt++]=(i-tmp-1); tmp=i; } } // FOR(i,0,cnt){ // PD(dis[i]);LN; // } sort(dis,dis+cnt+1); int count=1; int mdis=S-(mm-1)-(S-mx); // PD(mdis);LN; while(count<M){ mdis-=dis[cnt]; count++; cnt--; if(cnt<0) break; } PD(mdis);LN; return 0; }