CodeForces 566B Replicating Processes

 #include <bits/stdc++.h>
 #define N 3010
 #define LL long long
 #define unsigned U
 using namespace std;
 int cas=,T;
 int n,a[N<<],b[N<<],c[N<<],v[N],s[N],vis[N<<];
 int main()
 {
     //freopen("1.in","w",stdout);
     //freopen("1.in","r",stdin);
     //freopen("1.out","w",stdout);
     //scanf("%d",&T);
     while(scanf("%d",&n)==)
     {
         memset(v,,sizeof(v));
         memset(vis,,sizeof(vis));
         for(int i=;i<=n<<;i++) scanf("%d%d%d",a+i,b+i,c+i);
         int all=n<<;
         for(int i=;i<=n;i++) s[i]=;
         puts("YES");
         while(all)
         {
             for(int i=;i<=n<<;i++)
             {
                 if(!vis[i])
                 {
                     //for(int j=1;j<=n;j++) printf("\n%d %d\n",s[j],v[j]);
                     if(b[i]==c[i]&&s[b[i]]+v[b[i]]<)
                     {
                         vis[i]=;
                         all--;
                         s[a[i]]--;
                         v[b[i]]+=;
                         printf("%d ",i);
                         //break;
                     }
                     else if(b[i]!=c[i]&&s[b[i]]+v[b[i]]<&&s[c[i]]+v[c[i]]<)
                     {
                         vis[i]=;
                         all--;
                         s[a[i]]--;
                         v[b[i]]++;
                         v[c[i]]++;
                         printf("%d ",i);
                         //break;
                     }
                 }
             }
         }
         printf("\n");
     }
     //printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
     return ;
 }

第二次打的代码

 #include <stdio.h>
 #include <cstring>
 #include <iostream>
 #include <iterator>

 using namespace std;

 const int maxn = ;
 int a[ * maxn + ], b[ * maxn + ], c[ * maxn + ];
 int ans[ * maxn + ], vis[ * maxn + ];
 int cnt[maxn + ], m;

 void f(int n) {
     m = ;
     while (true) {
         bool flag = true;
         for (int i = ; i <=  * n; ++i) {
             if (vis[i]) continue;
             if ((b[i] == c[i] and cnt[b[i]] +  <= )
                 or (b[i] != c[i] and cnt[b[i]] +  <=  and cnt[c[i]] +  <= )) {
                 --cnt[a[i]];
                 ++cnt[b[i]];
                 ++cnt[c[i]];
                 ans[m++] = i;
                 vis[i] = ;
                 flag = false;
             }
         }
         if (flag)
             return;
     }
 }

 int main() {
     cin.tie();
     ios::sync_with_stdio(false);

     int n;
     while (scanf("%d",&n)==) {
         memset(vis, , sizeof(vis));
         memset(cnt, , sizeof(cnt));
         for (int i = ; i <=  * n; ++i) scanf("%d%d%d",a+i,b+i,c+i);
         f(n);

         if (m !=  * n)
             cout << "NO" << endl;
         else {
             cout << "YES" << endl;
             copy(ans, ans +  * n, ostream_iterator<int>(cout, " "));
             cout << endl;
         }
     }
     return ;
 }

师兄代码

　　为减轻服务器负担，将一个进程分为两个进程，每个服务器上原来有四个进程，分完后每个服务器有八个进程，分的过程中每个服务器最多有九个进程

　　做法：先把每个服务器的四个小进程放到同一个数组，循环遍历所有服务器，遍历到每个服务器时就将其中一个服务器分成两个，如果不符合分解条件就先不分，分下一个，一直循环到所有服务器原来的四个小进程都分完

 #include<cstdio>
 #include<cstring>
 #include<vector>
 using namespace std;
 struct node
 {
     int b, c, i;
     node(int x = , int y = , int z = )
     {
         b = x;
         c = y;
         i = z;
     }
 };
 vector<node>a[];
 int n;
 char vis[];
 int main()
 {
     while (scanf("%d", &n) == )
     {
         memset(vis, , sizeof(vis));
         int i, tmp, tmp1, tmp2, j,num;
         for (i = ; i <= n; i++)
             a[i].clear();
         n *= ;
         num = n;
         for (i = ; i <= n; i++)
         {
             scanf("%d%d%d", &tmp, &tmp1, &tmp2);
             a[tmp].push_back(node(tmp1, tmp2, i));
         }
         n /= ;
         puts("YES");
         int work = ;
         while (work)
         {
             work = ;
             for (j = ; j <= n; j++)
             {
                 if (!a[j].empty())
                     for (vector<node>::iterator k = a[j].begin(); k != a[j].end(); k++)
                     {
                         --vis[j];
                         ++vis[k->b];
                         ++vis[k->c];
                         work = ;
                         if (vis[k->b] <  && vis[k->c] < )
                         {
                             printf("%d", k->i);
                             num--;
                             if (num) printf(" ");
                             a[j].erase(k);
                             break;
                         }
                         else
                         {
                             ++vis[j];
                             --vis[k->b];
                             --vis[k->c];
                         }
                     }
             }
         }
         puts("");
     }
     return ;
 }