hdu 3572 Task Schedule 最大流 Dinic算法，，卡时间。。建图非常有讲究

Task Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4617    Accepted Submission(s): 1513

Problem Description

Our geometry princess XMM has stoped her study in computational geometry to concentrate on her newly opened factory. Her factory has introduced M new machines in order to process the coming N tasks. For the i-th task, the factory has to start processing it at or after day Si, process it for Pi days, and finish the task before or at day Ei. A machine can only work on one task at a time, and each task can be processed by at most one machine at a time. However, a task can be interrupted and processed on different machines on different days.
Now she wonders whether he has a feasible schedule to finish all the tasks in time. She turns to you for help.

 

Input

On the first line comes an integer T(T<=20), indicating the number of test cases.

You are given two integer N(N<=500) and M(M<=200) on the first line of each test case. Then on each of next N lines are three integers Pi, Si and Ei (1<=Pi, Si, Ei<=500), which have the meaning described in the description. It is guaranteed that in a feasible schedule every task that can be finished will be done before or at its end day.

 

Output

For each test case, print “Case x: ” first, where x is the case number. If there exists a feasible schedule to finish all the tasks, print “Yes”, otherwise print “No”.

Print a blank line after each test case.

 

Sample Input

24 31 3 5 1 1 42 3 73 5 92 22 1 31 2 2

 

Sample Output

Case 1: Yes   Case 2: Yes

 最大流。这个题的建图算是经典，因为限定每个时刻每台机器只能处理一个任务，所以可以把时间点分配给各个合法的机器...具体是先设定一个超级源点S，连向各个任务，容量为该任务所需时间，各个任务连向在范围内的时间点，容量为1（保证每个时刻xxx这个条件），所有时间点连向超级汇点T，容量为机器台数，最后求最大流，等于所有机器所需时间和的就是yes

代码：

#include <cstdio>#include <cstring>#include <queue>#define SIZE 1100#define INF 1000000000using namespace std ;struct Edge{int to , w , next ;}edge[SIZE*SIZE];int n , m , start , ends , index = 0;int  head[SIZE] , level[SIZE] , cur[SIZE];bool visited[SIZE] ;bool bfs(){queue<int> que ;que.push(start) ;memset(level,-1,sizeof(level)) ;level[start] = 0 ;while(!que.empty()){int pos = que.front() ;que.pop() ;for(int next = head[pos] ; next != -1 ; next = edge[next].next){if(level[edge[next].to]<0 && edge[next].w>0){level[edge[next].to] = level[pos]+1 ;que.push(edge[next].to) ;}}}return level[ends] != -1 ;}int min(int a , int b){return a>b?b:a ;}int dfs(int pos , int flow){int deta = 0 , tmp = 0;if(pos == ends)return flow ;for(int next = head[pos] ; next != -1 ; next = edge[next].next){if(edge[next].w > 0 && level[pos] == level[edge[next].to]-1){tmp = dfs(edge[next].to,min(flow-deta,edge[next].w)) ;if(tmp>0){edge[next].w -= tmp ;edge[next^1].w += tmp ;deta += tmp ;if(deta == flow)break ;}elselevel[edge[next].to] = -1 ;//不加这个，，超时。。。 }}return deta ;}int dinic(){int ans = 0 , flow = 0 ;while(bfs()){int deta = 0 ;ans += dfs(0,INF) ;}return ans ;}void add(int s , int d , int w){edge[index].next = head[s] ;edge[index].to = d ;edge[index].w = w ;head[s] = index ++ ;edge[index].next = head[d] ;edge[index].to = s ;edge[index].w = 0 ;head[d] = index ++ ;}int main(){int t , c = 1 ;scanf("%d",&t) ;while(t--){scanf("%d%d",&n,&m) ;memset(head,-1,sizeof(head)) ;index = 0 ;int sum = 0 , max = -1;start = 0;for(int i = 1 ; i <= n ; ++i){int x , y , z ;scanf("%d%d%d",&x,&y,&z) ;sum += x ;max = max>z?max:z ;add(start,i,x) ;for(int j = y ; j <= z ; ++j){add(i,j+n,1) ;}}ends = n+max+1 ;for(int i = 1 ; i <= max ; ++i)add(i+n,ends,m) ;int ans = dinic() ;printf("Case %d: ",c++) ;if(ans != sum)puts("No\n") ;elseputs("Yes\n");}return 0 ;}

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