【杭电2647】Reward

Reward

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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 11 22 21 22 1

Sample Output

1777-1

算是copy别人的。。本渣渣没用过vector做邻链表

#include <stdio.h>  #include <string.h>  #include <vector>  using namespace std;    int n;  int indegree[10005];  vector <int>  v[10005];  //用vector做邻链表  //拓扑排序 int topsort(){      int i,j,tag;      int countn = 0,q = 0;      int tem[10005];      int sum = n * 888;  //每个人底钱是888       while(countn != n){          tag = 0;          for(i = 1;i <= n;i++){              if(indegree[i]==0)//入度为0 {                  indegree[i] = -1;  //变为-1，防止下次被找到                 tem[tag] = i; //排过序的点存入数组                 tag++;  //记录排过点的个数             }          }          if(tag==0)//没有入度为0的点，可能存在双向的 {              return -1;          }          else{              sum += q * tag;              q++;              countn += tag;              for(i = 0;i < tag;i++){                  for(j = 0;j < v[tem[i]].size();j++){                      indegree[v[tem[i]].at(j)]--;                  }              }          }      }      return sum;  }  int main(){      int m;      int i,j;      int a,b;        while(scanf("%d%d",&n,&m) != EOF){          for(i = 0;i < 10005;i++)              v[i].clear();          memset(indegree,0,sizeof(indegree));          while(m--){              scanf("%d%d",&a,&b);              int flag = 0;              for(i = 0;i < v[b].size();i++){                  if(v[b].at(i) == a)                      flag = 1;              }              if(!flag){                  indegree[a]++;                  v[b].push_back(a);              }          }          printf("%d\n",topsort());      }      return 0;  }  

队列存储：

#include<cstdio>#include<cstring>#include<queue>using namespace std;int head[10010];int indegree[10010];int mon[10010];int num,sum;struct stu{   int to,next;}edge[20010];void inin(){//初始化     memset(indegree,0,sizeof(indegree)); memset(head,-1,sizeof(head));memset(mon,0,sizeof(mon));num=sum=0;}void add(int a,int b){//添加边    stu E={b,head[a]};   edge[num]=E;   head[a]=num++;   indegree[b]++;}void topo(int n){int i,j,id,t,ans=0;    queue<int>q;    while(!q.empty()) q.pop();    for(i=1;i<=n;i++){        if(indegree[i]==0){             q.push(i);             mon[i]=888;//第一个最少为888 }    }    while(!q.empty()){    t=q.front();    q.pop();    indegree[t]=-1;    sum+=mon[t];    ans++;//记录排过序的个数     for(i=head[t];i!=-1;i=edge[i].next){    id=edge[i].to;    if(--indegree[id]==0){    mon[id]=mon[t]+1;//每次比前一个多一     q.push(id);    }    }    }    if(ans==n) printf("%d\n",sum);    else printf("-1\n");}int main(){int n,m,i,j,a,b;while(scanf("%d%d",&n,&m)!=EOF){inin();for(i=1;i<=m;i++){   scanf("%d%d",&a,&b);   add(b,a);}topo(n);}return 0;}