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zc
2019-12-21 19:47:00
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f[x]表示状态为x时最少走多少步

枚举现在取哪两个点,然后

f[x|2^{i-1}|2^{j-1}]=\min(f[x|2^{i-1}|2^{j-1}],f[x])

如果直接枚举不剪枝的话可能会TLE

所以我们只要找到第一位是1的,然后往后面枚举就可以了,因为不关怎么样我们都要把全部取完,先取后取是一样的

(如果不懂可以自己模拟一下)

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i=x;i<=y;i++)
    #define Fdr(i,x,y) for(int i=x;i>=y;i--)
    #define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 0x3f3f3f3f
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const char* ln="\n";const int str=1<<20;struct IN{char buf[str],*s,*t;bool _;IN():s(buf),t(buf),_(0){}il char gc(){return s==t&&((t=(s=buf)+fread(buf,1,str,stdin))==s)?EOF:(*s++);}IN&operator>>(char&ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)_=1;else ch=c;return *this;}IN& operator>>(char* ch){clr(ch,0);if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)_=1;return *this;}IN& operator>>(string& ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)_=1;return *this;}template<typename T>IN&operator>>(T&x){if(_)return *this;char c=gc();bool ff=0;while(c!=EOF&&(c<'0'||c>'9'))ff^=(c=='-'),c=gc();if(c==EOF){_=1;return *this;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=(x<<3)+(x<<1)+c-48,c=gc();if(c==EOF)_=1;if(ff)x=-x;return *this;}}in;struct OUT{char buf[str],*s,*t;OUT():s(buf),t(buf+str){}~OUT(){fwrite(buf,1,s-buf,stdout);}void pt(char c){(s==t)?(fwrite(s=buf,1,str,stdout),*s++=c):(*s++=c);}OUT&operator<<(const char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(string s){for(int i=0;s[i];i++)pt(s[i]);return *this;}template<typename T>OUT&operator<<(T x){if(!x)return pt('0'),*this;if(x<0)pt('-'),x=-x;char a[30],t=0;while(x)a[t++]=x%10,x/=10;while(t--)pt(a[t]+'0');return *this;}}out;}using namespace IO;
#define N 10001
int n;
struct node{
    int x,y;
}a[N];
int f[1<<24];
int d(int i,int j){
    return (a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y);
}
int calc(int i,int j){
    return MIN(d(0,i)+d(i,j)+d(j,0),d(0,j)+d(j,i)+d(i,0));
}
int main(){
    cin>>a[0].x>>a[0].y>>n;
    Fur(i,1,n)cin>>a[i].x>>a[i].y;
    out<<"23333\n";
    clr(f,126);
    f[0]=0;
    Fur(x,0,(1<<24)-1)
        Fur(i,1,24)
            Fur(j,i+1,24)
            if(!(x&(1<<(i-1)))&&!(x&(1<<(j-1)))){
                int y=(x|(1<<(i-1)))|(1<<(j-1));
                f[y]=MIN(f[y],f[x]+calc(i,j));
            }
    out<<f[(1<<24)-1]<<ln;
}
LG CF8C Looking-for-Order
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