HGOI 20181028 题解

HGOI 20181028（复赛备考）

/*
真是暴力的一天，最后一题MLE？由于数组开得太大了！！！
270滚粗
考场上好像智商高了很多？！(假的)
*/

sol：暴力求解，然后没有数据范围吐槽一下（我开了10000000）

code：（100pts）

# include <bits/stdc++.h>
using namespace std;
const int MAXN=1e7+10;
char s[MAXN];
int fun(char c)
{
    if (c=='W') return 64;if (c=='H') return 32;if (c=='Q') return 16;
    if (c=='E') return 8; if (c=='S') return 4; if (c=='T') return 2;
    if (c=='X') return 1;
}
bool check(int l,int r)
{
    int ret=0;
    for (int i=l;i<=r;i++) ret+=fun(s[i]);
    if (ret==64) return true;
    else return false;
}
int main()
{
    freopen("jingle.in","r",stdin);
    freopen("jingle.out","w",stdout);
    cin>>s; int len=strlen(s);
    int cnt=0; for (int i=0;i<len;i++) if (s[i]=='/') cnt++;
    cnt--;
    int l=1,r,now=1;
    int ans=0;
    while(true) {
        for (r=l+1;r<=len;r++) if (s[r]=='/') break; r--;
        if (check(l,r)) ans++;
        l=r+2;
        if (now==cnt) break;
        now++;
    }
    printf("%d
",ans);
    return 0;
}

其实看一下就可以发现奇环显然是不行的。偶环一定可以通过0和1求解，然后就想到二分图

显然，如果这是张二分图那么就Yes采取01染色法求解（dfs暴力O(n)），如果不能做到01染色那么就输出No

但是注意图并不一定是连通图所以要多遍dfs

code(100pts)——第一次在考场上写read和write很激动然后封装了Input Output Base System的struct！码风很奇怪。。。

# include <bits/stdc++.h>
# define Rint register int
using namespace std;
const int MAXN=10005,MAXM=2*100005;
int head[MAXN],col[MAXN],tot=0,n,m;
bool vis[MAXN],ff;
struct rec{ int pre,to;}a[MAXM];
struct IOBS{
    inline int read()
    {
        int X=0,w=0; char c=0;
        while (!(c>='0'&&c<='9')) { w|=c=='-';c=getchar();}
        while (c>='0'&&c<='9') { X=(X<<3)+(X<<1)+(c^48); c=getchar();}
        return w?-X:X;
    }
    inline void write(Rint x)
    {
        if (x<0) { x=-x; putchar('-');}
        if (x>9) write(x/10);
        putchar(x%10+'0');
    }
    inline void write_files(Rint x,char cc){ write(x); putchar(cc);}
    inline void Files()    { freopen("perfect.in","r",stdin); freopen("perfect.out","w",stdout);}
}IO;
inline void adde(Rint u,Rint v)
{
    a[++tot].pre=head[u];
    a[tot].to=v;
    head[u]=tot;
}
inline void dfs(Rint u,Rint c)
{
    vis[u]=true; col[u]=c;
    for (Rint i=head[u];i;i=a[i].pre){
        int v=a[i].to; 
        if (vis[v]&&col[v]!=(!c)) { ff=false; return;}
        if (vis[v]&&col[v]==(!c)) continue;
        dfs(v,!c);
    }
}
int main()
{
    IO.Files(); n=IO.read(); m=IO.read();
    int u,v; 
    for (Rint i=1;i<=m;i++) {
        u=IO.read();v=IO.read();
        adde(u,v); adde(v,u);
    }
    memset(vis,false,sizeof(vis));
    for (Rint i=1;i<=n;i++) {
        if (vis[i]) continue;
        ff=true; dfs(i,0);
        if (ff==false) { printf("NO
"); return 0;}
    }
    putchar('Y');putchar('E');putchar('S');putchar('
');
    for (Rint i=1;i<=n;i++) 
    if (i!=n) IO.write_files(col[i],' ');
    else IO.write(col[i]);
    putchar('
');
    return 0;
}

 

sol： 通过n,m<=5发现是暴力题然后算了下裸bfs暴力会TLE但是还是打了23333！

我可是码过斗地主、德州扑克的人！其实这点码量差不多就是100行左右吧

不大不大 bfs套bfs！ （注意Mle！）

# include <bits/stdc++.h>
# define Rint register int
using namespace std;
const int MAXN=7;
const int dx[]={0,-1,0,1,0};
const int dy[]={0,0,1,0,-1};
struct node{ int M[MAXN][MAXN],L,scr;};
struct rec{ int x,y,step;};
char s[MAXN];
int mp[MAXN][MAXN],start[MAXN][MAXN],n,m;
bool inq[MAXN][MAXN];
queue<rec>Q;  
inline bool check(Rint X1,Rint Y1,Rint X2,Rint Y2,int &d)
{
    while(!Q.empty()) Q.pop(); 
    memset(inq,false,sizeof(inq));
    inq[X1][Y1]=true; rec st; st.step=0; st.x=X1; st.y=Y1;
    Q.push(st);
    while (!Q.empty()) {
        rec u=Q.front();Q.pop(); 
        for (int i=1;i<=4;i++) {
            rec v;
            v.x=u.x+dx[i]; v.y=u.y+dy[i]; v.step=u.step+1;
            if (v.x==X2&&v.y==Y2) { d=v.step-1;  return true;}
            if (v.x>n||v.x<1|v.y>m||v.y<1||inq[v.x][v.y]||mp[v.x][v.y]!=10) continue;
            if (v.x==X2&&v.y==Y2) { d=v.step-1;  return true;}
            Q.push(v); inq[v.x][v.y]=true;
        }
    }
    return false;
}
queue<node>q;
inline void bfs()
{
    node st; memcpy(st.M,start,sizeof(start));
    st.L=0; st.scr=0;
    int ans_scr=0,ans_L=0x7f7f7f7f;
    q.push(st);
    while (!q.empty()) {
        node u=q.front();q.pop();
        for (Rint X1=1;X1<=n;X1++)
        for (Rint Y1=1;Y1<=m;Y1++) {
          if (u.M[X1][Y1]==-1||u.M[X1][Y1]==10) continue;
             for (Rint X2=1;X2<=n;X2++)
             for (Rint Y2=1;Y2<=m;Y2++) {
                  if (X1==X2&&Y1==Y2) continue;
                  if (u.M[X2][Y2]==-1||u.M[X2][Y2]==10) continue;
                  if (u.M[X1][Y1]!=u.M[X2][Y2]) continue;
                  int delt_L; memcpy(mp,u.M,sizeof(u.M));
                  if (check(X1,Y1,X2,Y2,delt_L)) {
                      node v=u; memcpy(v.M,u.M,sizeof(u.M));
                      v.L=u.L+delt_L; v.scr=u.scr+1;
                      v.M[X1][Y1]=v.M[X2][Y2]=10;
                      if (v.scr>ans_scr) { ans_scr=v.scr; ans_L=v.L; }
                      else if (v.scr==ans_scr&&v.L<ans_L) ans_L=v.L; 
                      q.push(v);
                  }
              }
         }
    }
    if (ans_scr==0) printf("0 0
"); 
    else printf("%d %d
",ans_scr,ans_L);
}
int main()
{
    scanf("%d%d",&n,&m);
    for (Rint i=1;i<=n;i++) {
        cin>>s;
        for (Rint j=0;j<m;j++)
        if (s[j]>='0'&&s[j]<='9') start[i][j+1]=s[j]-'0';
        else if (s[j]=='X') start[i][j+1]=-1;
    }
    bfs();
    return 0;
}