CodeForces 566B Replicating Processes

#include <bits/stdc++.h>
#define N 3010
#define LL long long
#define unsigned U
using namespace std;
int cas=1,T;
int n,a[N<<2],b[N<<2],c[N<<2],v[N],s[N],vis[N<<2];
int main()
{
    //freopen("1.in","w",stdout);
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    //scanf("%d",&T);
    while(scanf("%d",&n)==1)
    {
        memset(v,0,sizeof(v));
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n<<2;i++) scanf("%d%d%d",a+i,b+i,c+i);
        int all=n<<2;
        for(int i=1;i<=n;i++) s[i]=4;
        puts("YES");
        while(all)
        {
            for(int i=1;i<=n<<2;i++)
            {
                if(!vis[i])
                {
                    //for(int j=1;j<=n;j++) printf("
%d %d
",s[j],v[j]);
                    if(b[i]==c[i]&&s[b[i]]+v[b[i]]<8)
                    {
                        vis[i]=1;
                        all--;
                        s[a[i]]--;
                        v[b[i]]+=2;
                        printf("%d ",i);
                        //break;
                    }
                    else if(b[i]!=c[i]&&s[b[i]]+v[b[i]]<9&&s[c[i]]+v[c[i]]<9)
                    {
                        vis[i]=1;
                        all--;
                        s[a[i]]--;
                        v[b[i]]++;
                        v[c[i]]++;
                        printf("%d ",i);
                        //break;
                    }
                }
            }
        }
        printf("
");
    }
    //printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
    return 0;
}

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

using namespace std;

const int maxn = 30000;
int a[4 * maxn + 10], b[4 * maxn + 10], c[4 * maxn + 10];
int ans[4 * maxn + 10], vis[4 * maxn + 10];
int cnt[maxn + 10], m;

void f(int n) {
    m = 0;
    while (true) {
        bool flag = true;
        for (int i = 1; i <= 4 * n; ++i) {
            if (vis[i]) continue;
            if ((b[i] == c[i] and cnt[b[i]] + 2 <= 5)
                or (b[i] != c[i] and cnt[b[i]] + 1 <= 5 and cnt[c[i]] + 1 <= 5)) {
                --cnt[a[i]];
                ++cnt[b[i]];
                ++cnt[c[i]];
                ans[m++] = i;
                vis[i] = 1;
                flag = false;
            }
        }
        if (flag)
            return;
    }
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n;
    while (scanf("%d",&n)==1) {
        memset(vis, 0, sizeof(vis));
        memset(cnt, 0, sizeof(cnt));
        for (int i = 1; i <= 4 * n; ++i) scanf("%d%d%d",a+i,b+i,c+i);
        f(n);

        if (m != 4 * n)
            cout << "NO" << endl;
        else {
            cout << "YES" << endl;
            copy(ans, ans + 4 * n, ostream_iterator<int>(cout, " "));
            cout << endl;
        }
    }
    return 0;
}

　　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 = 0, int y = 0, int z = 0)
    {
        b = x;
        c = y;
        i = z;
    }
};
vector<node>a[30010];
int n;
char vis[30010];
int main()
{
    while (scanf("%d", &n) == 1)
    {
        memset(vis, 4, sizeof(vis));
        int i, tmp, tmp1, tmp2, j,num;
        for (i = 0; i <= n; i++)
            a[i].clear();
        n *= 4;
        num = n;
        for (i = 1; i <= n; i++)
        {
            scanf("%d%d%d", &tmp, &tmp1, &tmp2);
            a[tmp].push_back(node(tmp1, tmp2, i));
        }
        n /= 4;
        puts("YES");
        int work = 1;
        while (work)
        {
            work = 0;
            for (j = 1; 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 = 1;
                        if (vis[k->b] < 10 && vis[k->c] < 10)
                        {
                            printf("%d", k->i);
                            num--;
                            if (num) printf(" ");
                            a[j].erase(k);
                            break;
                        }
                        else
                        {
                            ++vis[j];
                            --vis[k->b];
                            --vis[k->c];
                        }                    
                    }
            }
        }
        puts("");
    }
    return 0;
}