2014 牡丹江区域赛 B D I

http://acm.zju.edu.cn/onlinejudge/showContestProblems.do?contestId=358

The 2014 ACM-ICPC Asia Mudanjiang Regional Contest

136 - The 2014 ACM-ICPC Asia Mudanjiang Regional Contest

Solved	ID	Title	Ratio (AC/All)
Yes	A	Average Score	61.78% (456/738)
Yes	B	Building Fire Stations	10.40% (36/346)
	C	Card Game	0.00% (0/0)
Yes	D	Domination	40.46% (123/304)
	E	Excavator Contest	5.12% (2/39)
	F	Fiber-optic Network	23.07% (3/13)
	G	Garden and Sprinklers	0.00% (0/41)
	H	Hierarchical Notation	9.37% (30/320)
Yes	I	Information Entropy	53.19% (366/688)
	J	Jacobi Pattern	0.00% (0/2)
	K	Known Notation	20.96% (96/458)

B Building Fire Stations

题意：给出一个树，找两个点建消防站，使离最近的消防站的距离最远的点的这个距离最小。点有20W个。

题解：这2个消防站肯定在树的直径上，二分这个距离d，分别把2个消防站放在与直径的2个端点距离d的位置。

20W个点，不碉的深搜会Segment Fault，我的深搜可是加了const &的碉深搜，才能过的。

听说的更碉的题解：在直径的中点把树切成两半，分别求直径中点，就得了。

代码：

 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 using namespace std;
 #define ll long long
 #define usll unsigned ll
 #define mz(array) memset(array, 0, sizeof(array))
 #define mf1(array) memset(array, -1, sizeof(array))
 #define minf(array) memset(array, 0x3f, sizeof(array))
 #define REP(i,n) for(i=0;i<(n);i++)
 #define FOR(i,x,n) for(i=(x);i<=(n);i++)
 #define RD(x) scanf("%d",&x)
 #define RD2(x,y) scanf("%d%d",&x,&y)
 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
 #define WN(x) printf("%d\n",x);
 #define RE  freopen("D.in","r",stdin)
 #define WE  freopen("huzhi.txt","w",stdout)
 #define mp make_pair
 #define pb push_back
 #define pf push_front
 #define ppf pop_front
 #define ppb pop_back
 const double pi=acos(-1.0);
 const double eps=1e-;

 const int maxn=;
 const int maxm=*maxn;

 int n;

 struct Edge {
     int v,next;
 } e[maxm];
 int en,head[maxn];

 inline void add(const int &x,const int &y) {
     e[en].v=y;
     e[en].next=head[x];
     head[x]=en++;
 }

 int ans[];

 int anode,bnode;
 int len;
 inline void dfs1(const int &x,const int &f,const int &step) {
     if(step>len) {
         len=step;
         anode=x;
     }
     for(int i=head[x]; i!=-; i=e[i].next) {
         if(e[i].v != f) dfs1(e[i].v , x , step+);
     }
 }

 bool flag;
 int b[maxn],bn;
 inline void dfs2(const int &x,const int &f) {
     if(x==bnode) {
         flag=;
         b[bn++]=x;
         return;
     }
     for(int i=head[x]; i!=-; i=e[i].next) {
         if(e[i].v != f) {
             dfs2(e[i].v , x);
             if(flag) {
                 b[bn++]=x;
                 return;
             }
         }
     }
 }

 int _dis;
 int u[maxn];

 inline void dfs3(const int &x,const int &f,const int &step){
     if(step>_dis) {
         return;
     }
     u[x]=;
     for(int i=head[x]; i!=-; i=e[i].next) {
         if(e[i].v != f) dfs3(e[i].v , x , step+);
     }
 }

 inline bool ok(const int &dis){
     int i;
     int nodec=b[dis];
     int noded=b[bn--dis];
     len=;
     _dis=dis;
     memset(u,,sizeof(u));
     dfs3(nodec,-,);
     dfs3(noded,-,);
     FOR(i,,n)if(!u[i])return ;
     return ;
 }
 inline void farm() {
     int i;
     len=;
     dfs1(,-,);
     //printf("%d!\n",anode);
     bnode=anode;
     len=;
     dfs1(bnode,-,);
     //printf("%d!\n",anode);
     flag=,bn=;
     dfs2(anode,-);
 //    FOR(i,0,bn-1)printf("%d ,",b[i]);
 //    printf("(a=%d , b=%d)\n",anode,bnode);

     int an;
     int l=,r=bn-,mid;
     while(l<=r){
         mid=(l+r)>>;
 //        printf("%d,%d,%d\n",l,mid,r);
         if(ok(mid))r=mid-,an=mid;
         else l=mid+;
     }
     ans[]=an;
     ans[]=b[an];
     ans[]=b[bn--an];
     if(ans[]==ans[])ans[]=(ans[])%n+;
     return;
 }

 int main() {
     //re;
     int T;
     int i,j,x,y;
     scanf("%d",&T);
     while(T--) {
         scanf("%d",&n);
         memset(head,-,sizeof(head));
         en=;
         REP(i,n-) {
             scanf("%d%d",&x,&y);
             add(x,y);
             add(y,x);
         }
         farm();
         printf("%d %d %d\n",ans[],ans[],ans[]);
     }
     return ;
 }

D Domination

题意：在棋盘上没有棋的地方随机放棋，直到每一列每一行都有棋，求步数的期望。

题解：DP。dp[i][j][k]表示已经有i行有棋、j列有棋、已经放了k个棋 这种情况的概率。

我们是6000ms险过，也许DP的方式不对？

代码：

 #include <map>
 #include <set>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 #define REP(i,n) for(i=0;i<n;i++)
 #define FOR(i,x,y) for(i=x;i<=y;i++)
 #define mz(x) memset(x,0,sizeof(x))
 #define re freopen("in.txt","r",stdin)
 #define pb push_back
 #define ll long long
 #define maxn 50005
 #define mod 1000000009

 double dp[][][];
 int n,m;

 double dfs(int x,int y,int z) {
     if(x== || y== || z==)return dp[x][y][z]=0.0;
     if(dp[x][y][z]>=-0.5)return dp[x][y][z];
     dp[x][y][z]=dfs(x-,y,z-)*(n-x+)*y + dfs(x,y-,z-)*(m-y+)*x + dfs(x-,y-,z-)*(m-y+)*(n-x+);
     if(x!=n || y!=m) dp[x][y][z]+= dfs(x,y,z-)*(x*y-z+);
     dp[x][y][z]/=(double)(n*m-z+);
     //printf("(%d,%d,%d) %lf\n",x,y,z,dp[x][y][z]);
     return dp[x][y][z];
 }

 double farm(int _n,int _m) {
     int i,j,k;
     n=_n,m=_m;
     FOR(i,,n)FOR(j,,m)FOR(k,,n*m)dp[i][j][k]=-1.0;
     dp[][][]=1.0;
     FOR(i,,n)dp[i][][i]=dp[i-][][i-]*(n-i+)/(n*m-i+);
     FOR(i,,m)dp[][i][i]=dp[][i-][i-]*(m-i+)/(n*m-i+);
     double ans=0.0;
     FOR(i,,n*m) {
         ans+=dfs(n,m,i)*i;
         //printf("%d %lf\n",i,dfs(n,m,i));
     }
     //printf("%.12lf\n",E);
     return ans;
 }

 int main() {
     //re;
     int T;
     int n,m;
     scanf("%d",&T);
     while(T--) {
         scanf("%d%d",&n,&m);
         printf("%.12lf\n",farm(n,m));
     }
     return ;
 }

I Information Entropy

直接算，那啥是0的时候那个极限求完就是0。

代码：

 #include <map>
 #include <set>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 #define re freopen("in.txt","r",stdin)
 #define pb push_back
 #define ll long long
 #define maxn 50005
 #define mod 1000000009
 const double e=2.7182818284590452353602874;
 int n;
 int p[];
 char s[];

 double farm() {
     double hx=0.0;
     sort(p,p+n);
     if(s[]=='n') {
         for(int i=; i<n; i++) {
             if(p[i]!=) hx-=(1.0*p[i]) * (log(1.0*p[i]));
         }
         hx-=log(0.01)*100.0;
         return hx;
     } else if( s[]=='d') {
         for(int i=; i<n; i++) {
             if(p[i]!=) hx-=(1.0*p[i]) * (log10(1.0*p[i]));
         }
         hx+=200.0;
         return hx;
     } else if(s[]=='b') {
         for(int i=; i<n; i++) {
             if(p[i]!=) hx-=(1.0*p[i]) * (log2(1.0*p[i]));
         }
         hx-=log2(0.01)*100.0;
         return hx;
     }
 }

 int main() {
 //re;
     int T;
     scanf("%d",&T);
     while(T--) {
         scanf("%d%s",&n,s);
         for(int i=; i<n; i++) {
             scanf("%d",&p[i]);
         }
         printf("%.12lf\n",farm()/100.0);
     }
     return ;
 }