拓扑排序的变形，逆序建图就好了

Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3951    Accepted Submission(s): 1203

Problem Description Dandelion’s uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble about how to distribute the rewards.

The workers will compare their rewards ,and some one may have demands of the distributing of rewards ,just like a’s reward should more than b’s.Dandelion’s unclue wants to fulfill all the demands, of course ,he wants to use the least money.Every work’s reward will be at least 888 , because it’s a lucky number.  

Input One line with two integers n and m ,stands for the number of works and the number of demands .(n<=10000,m<=20000)

then m lines ,each line contains two integers a and b ,stands for a’s reward should be more than b’s.  

Output For every case ,print the least money dandelion ‘s uncle needs to distribute .If it’s impossible to fulfill all the works’ demands ,print -1.  

Sample Input

2 1 1 2 2 2 1 2 2 1  

Sample Output 普通拓扑排序

1777 -1

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
const int INF = 1e6;
const int ma = 1010;
#define MAX 0x3f3f3f3f
using namespace std;
struct node{
    int u,v,next;
}g[20100];
int head[20100],n,m,rudu[10100],t;
int cost[10100];
//struct no{
//    int x,ans;
//    friend bool operator < (const no &a,const no &b)
//    {
//        return a.ans>b.ans;
//    }
//};
void add(int a,int b)
{
    g[t].u = a;
    g[t].v = b;
    g[t].next = head[a];
    head[a] = t++;
}
int l;
bool vis[10010];
void TOP()
{
    //priority_queue<no>q;
    l = 0;
   // no f,t;
    memset(vis,0,sizeof(vis));
    for(int i = 1;i<=n;i++)
        cost[i] = 888;
        int sum = 0;
        int flag = 1;
        for(int i = 1;i<=n;i++)
        {
            for(int j = 1;j<=n;j++)
            {
                if(rudu[j]==0)
                {
                    rudu[j]--;
                    l++;
                    sum+=cost[j];
                    for(int k = head[j];k!=0;k = g[k].next)
                    {
                        rudu[g[k].v]--;
                        if(cost[g[k].v] < cost[j]+1) //这句不能少，要判断当前连接的点有没有被其他的点+1过，加过则不加
                        cost[g[k].v] = cost[j]+1;
                    }
                }
            }
        }
//printf("l = %d\n",l);
       if(l==n)
        printf("%d\n",sum);
       else
        puts("-1");

}
int main()
{
    int a,b;
    while(~scanf("%d%d",&n,&m))
    {
        memset(rudu,0,sizeof(rudu));
        memset(head,0,sizeof(head));
        t = 1;
        for(int i = 0;i<m;i++)
        {
             scanf("%d%d",&a,&b);
             add(b,a);
        rudu[a]++;
        }
        TOP();
    }
    return 0;
}

队列形式的拓扑排序

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
const int INF = 1e6;
const int ma = 1010;
#define MAX 0x3f3f3f3f
using namespace std;
struct node{
    int u,v,next;
}g[50000];
int head[50100],n,m,rudu[10100],t;
struct no{
    int x,ans;
};
void add(int a,int b)
{
    g[t].u = a;
    g[t].v = b;
    g[t].next = head[a];
    head[a] = t++;
}
int aa[1100],l;
void TOP()
{
    queue<no>q;
    l = 0;
    no f,t;
    t.ans = 888;
    bool vis[10010];
    memset(vis,0,sizeof(vis));
    long long sum = 0;
        for(int j = 1;j<=n;j++)
        {
            if(rudu[j]==0 && !vis[j])
            {
                rudu[j]--;
                vis[j] = 1;
                l++;
                t.x = j;
                q.push(t);
            }
        }
        while(!q.empty())
        {
            f = q.front();
            q.pop();
            sum += f.ans;
            if(!head[f.x]) continue;
            for(int i = head[f.x];i!=0;i = g[i].next)
            {
                rudu[g[i].v]--;
                if(rudu[g[i].v]==0 && !vis[g[i].v])
                {
                    l++;
                    if(t.ans<f.ans+1)
                    t.ans = f.ans+1;
                    t.x = g[i].v;
                    q.push(t);
                    vis[t.x] = 1;
                }
            }
        }
//printf("l = %d\n",l);
       if(l==n)
        cout<<sum<<endl;
       else
        puts("-1");

}
int main()
{
    int a,b;
    while(~scanf("%d%d",&n,&m))
    {
        memset(rudu,0,sizeof(rudu));
        memset(head,0,sizeof(head));
        t = 1;
        for(int i = 0;i<m;i++)
        {
             scanf("%d%d",&a,&b);
             add(b,a);
        rudu[a]++;
        }
           TOP();
    }
    return 0;
}