拓扑排序:

两个队列,一个放不需要重启入度为0的,一个放需要重启入度为0的....从不需要重启的队列开始,每弹出一个数就更新下入度,遇到入读为0的就加入到相应队列里,当队列空时,记录重启次数+1,交换队列..一直到两个队列都为空

Smart Software Installer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

The software installation is becoming more and more complex. An automatic tool is often useful to manage this process. An IT company is developing a system management utility to install a set of software packages automatically with the dependencies. They found
that reboot is often required to take effect after installing some software. A software package cannot be installed until all software packages it depends on are installed and take effect.

In the beginning, they implemented a simple installation algorithm, but the system would reboot many times during the installation process. This will have a great impact on the user experience. After some study, they think that this process can be further optimized
by means of installing as much packages as possible before each reboot.

Now, could you please design and implement this algorithm for them to minimize the number of restart during the entire installation process?

Input

The first line is an integer n (1 <= n <= 100), which is the number of test cases. The second line is blank. The input of two test cases is separated by a blank line.

Each test case contains m (1 <= n <= 1000) continuous lines and each line is no longer than 1024 characters. Each line starts with a package name and a comma (:). If an asterisk (*) exists between the package name and the comma, the reboot operation is required
for this package. The remaining line is the other package names it depends on, separated by whitespace. Empty means that there is no dependency for this software. For example, “a: b” means package b is required to be installed before package a. Package names
consist of letters, digits and underscores, excluding other special symbols.

Assume all packages here need to be installed and all referenced packages will be listed in an individual line to define the reboot property. It should be noted that cyclic dependencies are not allowed in this problem.

Output

For each test case, you should output a line starting with “Case #: " (# is the No. of the test case, starting from 1) and containing the reboot count for this test case. (Refer to the sample format)

Sample Input

2

glibc:
gcc*: glibc

uefi*:
gcc*:
raid_util*: uefi
gpu_driver*: uefi
opencl_sdk: gpu_drivergcc

Sample Output

Case 1: 1
Case 2: 2 

Source

2014上海全国邀请赛——题目重现（感谢上海大学提供题目）

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <string>
#include <sstream>
#include <map>
#include <queue>

using namespace std;

const int maxn=2000;

map<string,int> mSI;
int nm;
bool restart[maxn];
int degree[maxn];

int hash(string name)
{
    int ret=mSI[name];
    if(ret==0)
    {
        mSI[name]=nm++;
        ret=nm-1;
    }
    return ret;
}

struct Edge
{
    int to,next;
}edge[maxn*maxn];

int Adj[maxn],Size;

void init()
{
    memset(Adj,-1,sizeof(Adj)); Size=0;
}

void add_edge(int u,int v)
{
    edge[Size].to=v; edge[Size].next=Adj[u]; Adj[u]=Size++;
}

int TUOPU()
{
    queue<int> q[2];
    int a=0,b=1;
    int n=nm-1;
    for(int i=1;i<=n;i++)
    {
        if(degree[i]==0)
        {
            if(restart[i]==true) q[b].push(i);
            else q[a].push(i);
        }
    }
    int time=0;
    while(!q[a].empty()||!q[b].empty())
    {
        while(!q[a].empty())
        {
            int u=q[a].front(); q[a].pop();
            for(int i=Adj[u];~i;i=edge[i].next)
            {
                int v=edge[i].to;
                degree[v]--;
                if(degree[v]==0)
                {
                    if(restart[v]==true) q[b].push(v);
                    else q[a].push(v);
                }
            }
        }
        if(q[b].empty()==true) continue;
        time++;swap(a,b);
    }
    return time;
}

int main()
{
    int T_T,cas=1;
    scanf("%d",&T_T);
    getchar(); getchar();
    while(T_T--)
    {
        init();
        mSI.clear(); nm=1;
        memset(restart,false,sizeof(restart));
        memset(degree,0,sizeof(degree));

        string line,name;
        while(getline(cin,line))
        {
            if(line[0]==0) break;
            istringstream sin(line);
            sin>>name;
            bool flag=false;
            int sz=name.size();
            if(name[sz-2]=='*')
            {
                flag=true;
                name[sz-2]=0;
                name.resize(sz-2);
            }
            else
            {
                name[sz-1]=0;
                name.resize(sz-1);
            }

            int to=hash(name);
            restart[to]=flag;
            while(sin>>name)
            {
                int from=hash(name);
                add_edge(from,to);
                degree[to]++;
            }
        }
        printf("Case %d: %d\n",cas++,TUOPU());
    }
    return 0;
}