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  • 2017.10.20 模拟赛

    题目链接

    T1

    考虑遇到一个首位为奇数的233或者或者223

    会发生两者之间的相互转变

    那么只需要考虑剩余的操作数的奇偶性即可

    #include<cstring>
    #include<cstdio>
    #define N 1000010
    int n,k;
    char ch[N];
    
    int main()
    {
        freopen("trans.in","r",stdin);
        freopen("trans.out","w",stdout);
        while(scanf("%d%d",&n,&k)!=EOF)
        {
            scanf("%s",1+ch);
            int a=1;
            while(k--)
            {
                while((ch[a]!='2'||ch[a+1]!='3')&&a<=n) a++;
                if(a>n) break;
                if(a%2==1)
                {
                    if(ch[a+2]=='3') k%=2;
                    ch[a+1]='2';
                }
                else
                {
                    ch[a]='3';
                    a--;
                }
            }
            printf("%s
    ",1+ch);
        }
        fclose(stdin); fclose(stdout);
        return 0;
    }
    View Code

    T2

    把豆豆和砖块丢在一起dp
    f[i][j][k]:到第i行,蛇的长度还剩下j,从k位置出第i行的最大得分
    g[j][l][r]:蛇的长度剩下j,从当前行l到r区间内仍为死亡的最大得分
    初始化:对于每个i,g[j][k][k]=f[i-1][j-a[i][k]][k]+max(-a[i][k],0) 为初始状态
    方程g[j][l][r] = max(g[j-a[i][l]][l+1][r] + max(-a[i][j],0),g[j-a[i][r]][l][r-1]+max(-a[i][k],0));
    最后f[i][j][k] = max{g[j][l][r](1≤l≤k ≤r ≤5)}

    #include <cstdlib>
    #include <cstdio>
    #include <cmath>
    #define N 205
    
    bool fw[N][6],vis[N][15][2];
    int n,m,ans,fx[5]={0,0,1,-1},fy[5]={1,-1,0,0},a[N][6];
    inline int max(int a,int b) {return a>b?a:b;}
    void Dfs(int x,int y,int life,int point)
    {
        if(x==n+1)
        {
            ans=max(ans,point);
            return ;
        } 
        if(life+a[x][y]<0)
        {
            ans=max(ans,point);
            return;
        }
        fw[x][y]=1;
        for(int i=0;i<4;i++)
        {
            if(fx[i]==1&&vis[x][y][1]==1) continue;
            if(fy[i]==-1&&vis[x][y][0]==1) continue;
            int tx=x+fx[i],ty=y+fy[i];
            if(tx<=n+1&&tx>0&&ty<=5&&ty>0&&fw[tx][ty]==0)
            {
                if(a[x][y]>0) Dfs(tx,ty,life+a[x][y],point);
                else Dfs(tx,ty,life+a[x][y],point-a[x][y]);
            } 
        }
        fw[x][y]=0;
    }
    int main(int argc,char *argv[])
    {
        freopen("snakevsblock.in","r",stdin);
        freopen("snakevsblock.out","w",stdout);
        scanf("%d",&n);
        for(int i=1;i<=n;++i)
         for(int j=1;j<=5;++j)
          scanf("%d",&a[i][j]);
        scanf("%d",&m);
        for(int x,y,i=1;i<=m;++i)
         scanf("%d%d",&x,&y),vis[x][y][1]=1,vis[x][y+1][0]=1;
        Dfs(1,3,4,0);
        printf("%d
    ",ans);
        fclose(stdin); fclose(stdout);
        return 0;
    }
    考场10分代码
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    
    #define N 207
    #define M 10005
    
    using namespace std;
    int a[N][5],f[N][M][5],g[M][5][5];
    bool flag[N][4];
    int n,m,maxi,ans,val;
    
    inline int read()
    {
        int x=0,f=1;char c=getchar();
        while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();}
        while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
        return x*f;
    }
    
    int main()
    {
        freopen("snakevsblock.in","r",stdin);
        freopen("snakevsblock.out","w",stdout);
        n=read();int x,y;
        for(int i=1;i<=n;i++) for(int j=0;j<5;j++)
          a[i][j]=read();
        m=read();
        for(int i=1;i<=m;i++)
        {
            x=read();y=read();
            flag[x][y-1]=1;
        }
        memset(f,-0x7f7f7f,sizeof f);
        f[0][4][2]=0;maxi=n*50;
        
        for(int i=1;i<=n;i++)
        {
            memset(g,-0x7f7f7f,sizeof g);
            for(int j=0;j<=maxi;j++)
              for(int k=0;k<5;k++)
                if(j-a[i][k]>=0 && j-a[i][k]<=maxi)
                  f[i][j][k]=g[j][k][k]=f[i-1][j-a[i][k]][k]+max(-a[i][k],0);
    
            for(int l=1;l<=4;l++)
              for(int j=0,k=j+l;k<5;j++,k++)
                for(int v=0;v<=maxi;v++)
                {
                    if(!flag[i][j]  && (val=v-a[i][j])>=0 && val<=maxi) g[v][j][k] = g[val][j + 1][k] + max(-a[i][j], 0);//g[v][j][k]=max(g[v][j][k],g[val][j+1][k]+max(-a[i][j],0));
                    if(!flag[i][k-1]&& (val=v-a[i][k])>=0 && val<=maxi) g[v][j][k]=max(g[v][j][k],g[val][j][k-1]+max(-a[i][k],0));
                    for(int to=j;to<=k;to++) f[i][v][to]=max(f[i][v][to],g[v][j][k]);
                }
        }
        
        for(int i=0;i<=n;i++)
          for(int j=0;j<=maxi;j++)
            for(int k=0;k<5;k++)
              ans=max(f[i][j][k],ans);
              
        printf("%d
    ",ans);
        return 0;
    }
    std

    T3

    任意将一个站点设置为树的根,首先求出DFS序,倒序处理,对于每个站点,如果有以以其LCA的且未被切断的路径,则将这个站点标记为损坏,用树链剖分维护即可

    #include <cstdlib>
    #include <cstdio>
    #include <set>
    #define N 100005
    using namespace std;
    
    set<int>se;
    struct node
    {
        int x,y;
    }e[N],v[N*3];
    bool vis[N];
    int n,m,k,limit,which[N],fa[N];
    int find_(int x) {return x==fa[x]?x:fa[x]=find_(fa[x]);}
    bool judge()
    {
        for(int i=1;i<=n;++i) fa[i]=i;
        for(int i=1;i<=m;++i)
        {
            int x=e[i].x,y=e[i].y;
            if(se.find(x)==se.end()||se.find(y)==se.end()) continue; 
            int fx=find_(x),fy=find_(y);
            if(fx!=fy) fa[fy]=fx;
        }
        int sum=0;
        for(int i=1;i<=k;++i)
        {
            int x=v[i].x,y=v[i].y;
            if(x==y) continue; 
            int fx=find_(x),fy=find_(y);
            if(fx!=fy)  sum++;
        }
        if(sum) return true;
    }
    void Dfs(int goal,int pos)
    {
        if(se.size()==goal)
        {
            if(judge())
            {
                printf("%d
    ",goal);
                for(set<int> :: iterator it=se.begin();it!=se.end();++it)
                printf("%d ",*it);
                exit(0);
            }
            return ;
        }
        for(int i=pos+1;i<=limit;++i)
        {
            se.insert(which[i]);
            Dfs(goal,i);
            se.erase(which[i]);  
        }
    }
    int main(int argc,char *argv[])
    {
    //    freopen("ping.in","r",stdin);
    //    freopen("ping.out","w",stdout);
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;++i)
        {
            scanf("%d%d",&e[i].x,&e[i].y);
            int p=e[i].x>e[i].y?e[i].x:e[i].y;
        }
        scanf("%d",&k);
        for(int i=1;i<=k;++i)
        {
            scanf("%d%d",&v[i].x,&v[i].y);
            if(!vis[v[i].x]) vis[v[i].x]=true,which[++limit]=v[i].x;
            if(!vis[v[i].y]) vis[v[i].y]=true,which[++limit]=v[i].y;
        }
        for(int i=1;i<=limit;++i) Dfs(i,0);
        fclose(stdin);
        fclose(stdout);
        return 0;
    }
    考场挂掉的暴力
    #include <bits/stdc++.h>
    using namespace std;
    int n, m, P, u, v, to[200005], nxt[200005], p[100005], deep[100005], q[500005][2];
    int son[100005], fa[100005], size[100005], top[100005], dfsx[100005], cnt, ans;
    int lca[500005], Q[500005], lr[100005][2];
    int sta[100005];
    bool flag[400005];
    void dfs(int x)
    {
        size[x] = 1;
        for (int i = p[x]; i != -1; i = nxt[i])
            if (to[i] != fa[x])
            {
                fa[to[i]] = x;
                deep[to[i]] = deep[x] + 1;
                dfs(to[i]);
                if (son[x] == -1 || size[to[i]] > size[son[x]]) son[x] = to[i];
                size[x] += size[to[i]];
            }
    }
    void dfs1(int x)
    {
        dfsx[x] = ++cnt;
        if (son[x] != -1) top[son[x]] = top[x], dfs1(son[x]);
        for (int i = p[x]; i != -1; i = nxt[i])
            if (to[i] != fa[x] && to[i] != son[x])
                top[to[i]] = to[i], dfs1(to[i]);
    }
    int findlca(int x, int y)
    {
        while (1)
        {
            if (top[x] == top[y]) return deep[x] > deep[y]? y : x;
            if (deep[top[x]] > deep[top[y]]) x = fa[top[x]];
            else y = fa[top[y]];
        }
    }
    bool query(int x, int l, int r, int ll, int rr)
    {
        if (l == ll && r == rr) return flag[x];
        int mid = (l + r) >> 1, L = x << 1, R = L | 1;
        if (rr <= mid) return query(L, l, mid, ll, rr);
        else if (ll > mid) return query(R, mid + 1, r, ll, rr);
        else return query(L, l, mid, ll, mid) | query(R, mid + 1, r, mid + 1, rr);
    }
    void modify(int x, int l, int r, int to)
    {
        flag[x] = true;
        if (l == r) return;
        int mid = (l + r) >> 1, L = x << 1, R = L | 1;
        if (to <= mid) modify(L, l, mid, to);
        else modify(R, mid + 1, r, to);
    }
    bool Query(int x, int y)
    {
        while (1)
        {
            if (top[x] == top[y])
            {
                if (deep[x] < deep[y]) return query(1, 1, cnt, dfsx[x], dfsx[y]);
                else return query(1, 1, cnt, dfsx[y], dfsx[x]);
            }
            if (deep[top[x]] > deep[top[y]])
                if (query(1, 1, cnt, dfsx[top[x]], dfsx[x])) return true;
                else x = fa[top[x]];
            else
            {
                if (query(1, 1, cnt, dfsx[top[y]], dfsx[y])) return true;
                else y = fa[top[y]];
            }
        }
    }
    void work(int x)
    {
        for (int i = p[x]; i != -1; i = nxt[i])
            if (to[i] != fa[x])
                work(to[i]);
        for (int i = lr[x][0]; i <= lr[x][1]; i++)
            if (!Query(q[Q[i]][0], q[Q[i]][1]))
            {
                modify(1, 1, cnt, dfsx[x]);
                sta[++ans] = x;
                return;
            }
    }
    bool cmp(int x, int y) {return lca[x] < lca[y];}
    int main()
    {
        freopen("ping.in","r",stdin);
        freopen("ping.out","w",stdout);
        scanf("%d%d", &n, &m);
        for (int i = 0; i <= n; i++) p[i] = son[i] = -1, top[i] = size[i] = fa[i] = deep[i] = 0;
        for (int i = 1; i <= n * 4; i++) flag[i] = false;
        for (int i = 1; i <= m; i++)
        {
            scanf("%d%d", &u, &v);
            u--, v--;
            to[i * 2 - 1] = v;
            nxt[i * 2 - 1] = p[u];
            p[u] = i * 2 - 1;
            to[i * 2] = u;
            nxt[i * 2] = p[v];
            p[v] = i * 2;
        }
        deep[0] = 1;
        dfs(0);
        cnt = 0;
        dfs1(0);
        scanf("%d", &P);
        for (int i = 1; i <= P; i++)
        {
            scanf("%d%d", &u, &v);
            u--, v--;
            q[i][0] = u, q[i][1] = v;
            lca[i] = findlca(u, v);
            Q[i] = i;
        }
        sort(Q + 1, Q + P + 1, cmp);
        for (int i = 0; i <= n; i++) lr[i][0] = P + 1, lr[i][1] = 0;
        for (int i = 1; i <= P; i++)
        {
            lr[lca[Q[i]]][0] = min(lr[lca[Q[i]]][0], i);
            lr[lca[Q[i]]][1] = max(lr[lca[Q[i]]][1], i);
        }
        ans = 0;
        work(0);
        printf("%d
    ", ans);
        for (int i = 1; i <= ans; i++) printf("%d ", sta[i] + 1);
        return 0;
    }
    std
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  • 原文地址:https://www.cnblogs.com/ruojisun/p/7700125.html
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