1 #include <bits/stdc++.h> 2 #define N 3010 3 #define LL long long 4 #define unsigned U 5 using namespace std; 6 int cas=1,T; 7 int n,a[N<<2],b[N<<2],c[N<<2],v[N],s[N],vis[N<<2]; 8 int main() 9 { 10 //freopen("1.in","w",stdout); 11 //freopen("1.in","r",stdin); 12 //freopen("1.out","w",stdout); 13 //scanf("%d",&T); 14 while(scanf("%d",&n)==1) 15 { 16 memset(v,0,sizeof(v)); 17 memset(vis,0,sizeof(vis)); 18 for(int i=1;i<=n<<2;i++) scanf("%d%d%d",a+i,b+i,c+i); 19 int all=n<<2; 20 for(int i=1;i<=n;i++) s[i]=4; 21 puts("YES"); 22 while(all) 23 { 24 for(int i=1;i<=n<<2;i++) 25 { 26 if(!vis[i]) 27 { 28 //for(int j=1;j<=n;j++) printf(" %d %d ",s[j],v[j]); 29 if(b[i]==c[i]&&s[b[i]]+v[b[i]]<8) 30 { 31 vis[i]=1; 32 all--; 33 s[a[i]]--; 34 v[b[i]]+=2; 35 printf("%d ",i); 36 //break; 37 } 38 else if(b[i]!=c[i]&&s[b[i]]+v[b[i]]<9&&s[c[i]]+v[c[i]]<9) 39 { 40 vis[i]=1; 41 all--; 42 s[a[i]]--; 43 v[b[i]]++; 44 v[c[i]]++; 45 printf("%d ",i); 46 //break; 47 } 48 } 49 } 50 } 51 printf(" "); 52 } 53 //printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC); 54 return 0; 55 }
1 #include <stdio.h> 2 #include <cstring> 3 #include <iostream> 4 #include <iterator> 5 6 using namespace std; 7 8 const int maxn = 30000; 9 int a[4 * maxn + 10], b[4 * maxn + 10], c[4 * maxn + 10]; 10 int ans[4 * maxn + 10], vis[4 * maxn + 10]; 11 int cnt[maxn + 10], m; 12 13 void f(int n) { 14 m = 0; 15 while (true) { 16 bool flag = true; 17 for (int i = 1; i <= 4 * n; ++i) { 18 if (vis[i]) continue; 19 if ((b[i] == c[i] and cnt[b[i]] + 2 <= 5) 20 or (b[i] != c[i] and cnt[b[i]] + 1 <= 5 and cnt[c[i]] + 1 <= 5)) { 21 --cnt[a[i]]; 22 ++cnt[b[i]]; 23 ++cnt[c[i]]; 24 ans[m++] = i; 25 vis[i] = 1; 26 flag = false; 27 } 28 } 29 if (flag) 30 return; 31 } 32 } 33 34 int main() { 35 cin.tie(0); 36 ios::sync_with_stdio(false); 37 38 int n; 39 while (scanf("%d",&n)==1) { 40 memset(vis, 0, sizeof(vis)); 41 memset(cnt, 0, sizeof(cnt)); 42 for (int i = 1; i <= 4 * n; ++i) scanf("%d%d%d",a+i,b+i,c+i); 43 f(n); 44 45 if (m != 4 * n) 46 cout << "NO" << endl; 47 else { 48 cout << "YES" << endl; 49 copy(ans, ans + 4 * n, ostream_iterator<int>(cout, " ")); 50 cout << endl; 51 } 52 } 53 return 0; 54 }
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
1 #include<cstdio> 2 #include<cstring> 3 #include<vector> 4 using namespace std; 5 struct node 6 { 7 int b, c, i; 8 node(int x = 0, int y = 0, int z = 0) 9 { 10 b = x; 11 c = y; 12 i = z; 13 } 14 }; 15 vector<node>a[30010]; 16 int n; 17 char vis[30010]; 18 int main() 19 { 20 while (scanf("%d", &n) == 1) 21 { 22 memset(vis, 4, sizeof(vis)); 23 int i, tmp, tmp1, tmp2, j,num; 24 for (i = 0; i <= n; i++) 25 a[i].clear(); 26 n *= 4; 27 num = n; 28 for (i = 1; i <= n; i++) 29 { 30 scanf("%d%d%d", &tmp, &tmp1, &tmp2); 31 a[tmp].push_back(node(tmp1, tmp2, i)); 32 } 33 n /= 4; 34 puts("YES"); 35 int work = 1; 36 while (work) 37 { 38 work = 0; 39 for (j = 1; j <= n; j++) 40 { 41 if (!a[j].empty()) 42 for (vector<node>::iterator k = a[j].begin(); k != a[j].end(); k++) 43 { 44 --vis[j]; 45 ++vis[k->b]; 46 ++vis[k->c]; 47 work = 1; 48 if (vis[k->b] < 10 && vis[k->c] < 10) 49 { 50 printf("%d", k->i); 51 num--; 52 if (num) printf(" "); 53 a[j].erase(k); 54 break; 55 } 56 else 57 { 58 ++vis[j]; 59 --vis[k->b]; 60 --vis[k->c]; 61 } 62 } 63 } 64 } 65 puts(""); 66 } 67 return 0; 68 }