NOIP 2017 宝藏

题目

fi,jf_{i,j} 表示当前最大深度为i,已连通点为j的方案数。

转移的时候,枚举深度为i+1的点集合k,然后k中每个点都接在j中离它最近的点下面。

这个可以用计算disj,kdis_{j,k}来做。

但这样空间复杂度是O(4n)O(4^n)的。

显然jk=j\cap k=\varnothing,我们就可以把j集合标记为1,k集合标记为2,O(3n)O(3^n)的算。

那么总时间复杂度O(n3n)O(n3^n)

空间复杂度O(3n)O(3^n)

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/*
Author: CNYALI_LK
LANG: C++
PROG: 3959.cpp
Mail: cnyalilk@vip.qq.com
*/
#include<bits/stdc++.h>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
using namespace std;
const double eps=1e-8;
const double pi=acos(-1.0);
typedef long long ll;
typedef pair<int,int> pii;
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
#define min mmin
#define max mmax
#define abs aabs
int read(){
int s=0,base=1;
char c;
while(!isdigit(c=getchar()))if(c=='-')base=-base;
while(isdigit(c)){s=s*10+(c^48);c=getchar();}
return s*base;
}
char WriteIntBuffer[1024];
template<class T>void write(T a,char end){
int cnt=0,fu=1;
if(a<0){putchar('-');fu=-1;}
do{WriteIntBuffer[++cnt]=fu*(a%10)+'0';a/=10;}while(a);
while(cnt){putchar(WriteIntBuffer[cnt]);--cnt;}
putchar(end);
}
int f[15][4096];
int dis[531441],fa_Q[4096];
int a[12][12];
int stk[4096],t0p;
int lowbit(int x){return x&-x;}
int g(int a,int b){
return fa_Q[a]+fa_Q[b]*2;
}


int main(){
#ifdef cnyali_lk
freopen("3959.in","r",stdin);
freopen("3959.out","w",stdout);
#endif
int n,u,v,m;
n=read();
if(n==1){printf("0\n");return 0;}
m=read();
memset(f,0x3f,sizeof(f));
for(int i=0;i<n;++i){
f[1][1<<i]=0;
}
memset(dis,0x3f,sizeof(dis));

int t=1<<n;
for(int i=1;i<t;++i)fa_Q[i]=fa_Q[i>>1]*3+(i&1);
while(m){
u=1<<(read()-1);v=1<<(read()-1);
chkmin(dis[g(u,v)],read());
dis[g(v,u)]=dis[g(u,v)];
--m;
}

int _t=t-1;
for(int i=1;i<t;++i){
for(int j=i^_t;j;j=(j-1)&(i^_t))stk[++t0p]=j;
while(t0p){
if(stk[t0p]==lowbit(stk[t0p]))chkmin(dis[g(i,stk[t0p])],min(dis[g(i^lowbit(i),stk[t0p])],dis[g(lowbit(i),stk[t0p])]));
else chkmin(dis[g(i,stk[t0p])],dis[g(i,stk[t0p]^lowbit(stk[t0p]))]+dis[g(i,lowbit(stk[t0p]))]);
--t0p;
}
}
int ans=0x3f3f3f3f;
for(int i=1;i<n;++i){
for(int j=1;j<t;++j)for(int k=_t^j;k;k=(k-1)&(_t^j))if(dis[g(j,k)]!=0x3f3f3f3f)chkmin(f[i+1][j|k],f[i][j]+dis[g(j,k)]*i);
chkmin(ans,f[i][_t]);
}
printf("%d\n",ans);
return 0;
}
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