Reward
Reward
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3015 Accepted Submission(s): 907
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.
then m lines ,each line contains two integers a and b ,stands for a's reward should be more than b's.
map[MAX][MAX];
#include<stdio.h>
#include<string.h>
int degree[],map[][]; int vis[],pre[];
int main()
{
int n,m,a,b,i,j,k,t,p,flag,sum,temp;
while(~scanf("%d%d",&n,&m))
{
memset(degree,,sizeof(degree));
memset(map,,sizeof(map));
memset(vis,,sizeof(vis));
t = sum = flag = ;
while(m--)
{
scanf("%d%d",&a,&b);
if(!map[a][b])
{
degree[b]++;
map[a][b] = ;
}
}
for(i = ;i < n;i ++)
{
k = ;
for(j = ;j <= n;j ++)
{
if(degree[j] == && vis[j] == )
pre[k++] = j;
}
sum += (+t)*k;
for(j = ;j < k;j ++)
{
vis[pre[j]] = ;
for(p = ;p <= n;p ++)
{
if(map[pre[j]][p] == && vis[p]== )
{
degree[p]--;
map[pre[j]][p] = ;
}
}
}
t++;
}
for(i = ;i <=n;i ++)
{
if(degree[i])
{
flag = ;
break ;
}
}
if(!flag)
printf("%d\n",sum);
else
printf("-1\n");
}
return ;
}
下面是AC代码:
int head[MAX];
typedef struct
{
int to; //记录终点;
int next; //记录下一个节点的位置;
int ... //记录其他一些信息,与题有关;
}EdgeNode;
EdgeNode edge[MAX_m];
cin >> i >> j >> ...;
edge[k].to = j;
edge[k].next = head[i];
edge[k].... = ...;
head[i] = k++;
#include<stdio.h>
#include<string.h>
typedef struct
{
int to;
int next;
}EdgeNode;
EdgeNode Edge[];
int head[],node[];
int cnt,indegree[],vis[];
void init()
{
cnt = ;
memset(vis,,sizeof(vis));
memset(head,-,sizeof(head));
memset(indegree,,sizeof(indegree));
} void add_edge(int n,int m)
{
Edge[cnt].to = n;
Edge[cnt].next = head[m];
head[m] = cnt++;
} int main()
{
int n,m,a,b,i,j,k,t,p,sum;
while(~scanf("%d%d",&n,&m))
{
init();
sum = p = ;
while(m--)
{
scanf("%d%d",&a,&b);
indegree[a]++;
add_edge(a,b);
}
for(i = ;i < n;i ++)
{
t = ;
for(j = ;j <= n;j ++)
{
if(indegree[j] == && vis[j] == )
node[t++] = j;
}
sum += (+p)*t;
p++;
for(j = ;j < t;j ++)
{
vis[node[j]] = ;
for(k = head[node[j]];k != -;k = Edge[k].next)
{
if(!vis[Edge[k].to])
indegree[Edge[k].to]--;
}
}
}
for(i = ;i <= n;i ++)
{
if(indegree[i])
{
sum = -;
break ;
}
}
printf("%d\n",sum);
}
return ;
}
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