CF #328

A:每次在一列中遍历所有行即可，从上往下找找第一个能到顶端的”W”,从下往上照找第一个能到底端的“B”,最后比较较小的那个就行。但是注意题目有说“W”的先动，所以如果它们到达目的地的步数一样时，A获胜。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>#include<set>#include<bitset>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,num) scanf("%d%d%d",&a,&b,&num)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define ll long longconst double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;//#pragma comment(linker,"/STACK:1024000000,1024000000")int n,m;#define N 210#define M 100010#define Mod 1000000000#define p(x,y) make_pair(x,y)char mp[10][10];int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);//  freopen("out.txt", "w", stdout);#endif    while(sfs(mp[0])!=EOF){        for(int i=1;i<8;i++)            sfs(mp[i]);        int aans = INF;        int bans = -1;        for(int j=0;j<8;j++){            int x = INF,y = -1;            for(int i=0;i<8;i++){                if(mp[i][j] == 'B') break;                else if(mp[i][j] == 'W'){                    x = i;                    break;                }            }            for(int i=7;i>=0;i--){                if(mp[i][j] == 'W') break;                else if(mp[i][j] == 'B'){                    y = i;                    break;                }            }            aans = Min(aans,x);            bans = Max(bans,y);        }        bans = 7-bans;        if(aans <= bans)            printf("A\n");        else            printf("B\n");    }    return 0;}

B:找找规律，发现公式：(n-2)*(n-2)

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>#include<set>#include<bitset>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,num) scanf("%d%d%d",&a,&b,&num)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define ll long longconst double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;//#pragma comment(linker,"/STACK:1024000000,1024000000")ll n,m;#define N 210#define M 100010#define Mod 1000000000#define p(x,y) make_pair(x,y)int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);//  freopen("out.txt", "w", stdout);#endif    while(sfI(n)!=EOF){        pfI((n-2)*(n-2));    }    return 0;}

C:感觉这道题目思路不难，符合条件的就是这么几种数字：
1：[1,Min(t,Min(w,b))-1] 比如2，3，那么1就符合条件。
2：[lcm(a,b),lcm(a,b)+Min(w,b)-1],这里的lcm(a,b)代表的是a,b的公倍数，但注意有个特例就是最后一组在t范围内的上述区间，我们要判断t和lcm(a,b)+Min(w,b)-1的大小关系才行。
但是这道题目坑比较多，最明显的就是求公倍数时的技巧，我们不能直接求，因为数据过大，会超范围，解决方法就是用double类型存储求得的公倍数，这样就不会出现超范围的问题了。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>#include<set>#include<bitset>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,num) scanf("%d%d%d",&a,&b,&num)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define ll long longconst double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;//#pragma comment(linker,"/STACK:1024000000,1024000000")int n,m;#define N 210#define M 100010#define Mod 1000000000#define p(x,y) make_pair(x,y)int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);//  freopen("out.txt", "w", stdout);#endif    ll t,w,b;    while(scanf("%I64d%I64d%I64d",&t,&w,&b)!=EOF){            ll res=0;            res += Min(t,Min(w,b)-1); //注意点1，注意比较t和Min(w,b)-1的大小            ll ggcd = gcd(w,b);            double lm = (double)(w/ggcd)*b; //求公倍数时注意要用double来存储            ll num = (ll)(t/lm); //求得t区间内有几个公倍数区间         if(num>0){ //如果区间数大于0            res += num;            res += (num-1)*(Min(w,b)-1); //前num-1个区间            ll maxlcm = (ll)(lm*num);            ll tmp = Min(t-maxlcm,Min(w,b)-1); //最后一个区间得判断其和t的大小关系            res += tmp;          }          ll gd = gcd(res,t);          res /= gd;          t /= gd;          printf("%I64d/%I64d\n",res,t);    }    return 0;}

D：这是一道自己不太会的，学习了下，感觉也不难。首先，我们将所有要到达的点自己构造一颗最小虚树，使得这颗虚树包含所有我们要到达的点，求得这颗虚树的边长(即所有边的长度+1)edg，和这棵树的直径（即离得最远的两个点的距离）len，最后的答案就是edg*2-len.
求len的方法很简单，就是先用dfs求出一个端点，然后再用一次求出另外一个，它们之间的距离就是树的直径。
求edge的话就是求出这课虚树包含的点的个数，边长就是 点的个数-1。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>#include<set>#include<bitset>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,num) scanf("%d%d%d",&a,&b,&num)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define ll long longconst double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;//#pragma comment(linker,"/STACK:1024000000,1024000000")int n,m;#define N 210#define M 200010#define Mod 1000000000#define p(x,y) make_pair(x,y)vector<int> G[M];int vis[M],vis2[M];int nd1,nd2;int len;int edgnum;void dfs1(int x,int ds,int pre){ //x代表当前节点，ds代表直径，pre代表前导点    if(ds>len && vis[x]){       //直径变大了，且该点是要包含的        len = ds;        nd2 = x;    }    if(ds == len && vis[x] && nd2>x){       //直径一样，但是x的序号比较小(题目要求的就是最小的序号)        nd2 = x;    }    for(int i=0;i<(int)G[x].size();i++){    //遍历        if(G[x][i] == pre)            continue;        dfs1(G[x][i],ds+1,x);    }}int dfs2(int x,int pre){        //将该虚树要包含的点标记起来，vis2[x]如果不为0，那么x就是虚树上的一个点    if(vis[x])        vis2[x]++;    for(int i=0;i<(int)G[x].size();i++){        if(G[x][i] == pre)            continue;        vis2[x] += dfs2(G[x][i],x);    }    return vis2[x];}void dfs3(int x,int pre){       //统计该虚树中的所有点    if(vis2[x])        edgnum++;    for(int i=0;i<(int)G[x].size();i++){        if(G[x][i] == pre)            continue;        dfs3(G[x][i],x);    }}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);//  freopen("out.txt", "w", stdout);#endif    while(sfd(n,m)!=EOF){        for(int i=0;i<=n;i++) G[i].clear();        int u,v;        for(int i=1;i<n;i++){            sfd(u,v);            G[u].push_back(v);            G[v].push_back(u);        }        int x;        memset(vis,0,sizeof vis);        memset(vis2,0,sizeof vis2);        nd1 = INF;        for(int i=0;i<m;i++){            sf(x);            vis[x]=1;         //标记要包含的点            nd1 = Min(nd1,x);        }        if(m == 1){             //注意特判啊！！            printf("%d\n0\n",nd1);            continue;        }        len=-1;        dfs1(nd1,0,-1);     //先用dfs求出一个树的端点        nd1 = nd2;      //nd1为第一个端点        len = -1;        edgnum = 0;        dfs1(nd1,0,-1); //求出第二个端点,nd2就是        dfs2(nd2,-1);       //标记该虚树中的所有点        dfs3(nd2,-1);   //统计虚树的点的个数        printf("%d\n",Min(nd1,nd2));    //输出序号较小的        if(edgnum!=1)edgnum--;  //如果不只有1个点，那么边长 = 点的个数-1        printf("%d\n",edgnum*2-len);        //最后的答案就是边长*2-树的直径    }    return 0;}