Task Schedule
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3412 Accepted Submission(s): 1197
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
2
4 3
1 3 5
1 1 4
2 3 7
3 5 9
2 2
2 1 3
1 2 2
Sample Output
Case 1: Yes
Case 2: Yes
Author allenlowesy
思路：建一个超级源点0，然后假设工作区间长度为T ,再建立[1，T]个点，源点到每个点的流量为M（每天只有M台机器工作），接着，把相应的工作日向后平移T 天，每个工作日到相应的[1,T]的流量为1，到终点的流量也为1.
最后求最大流是否大于等于总总工作量就是了。
#include"stdio.h"
#include"string.h"
#include"queue"
using namespace std;
#define N 1005
#define max(a,b) (a>b?a:b)
#define min(a,b) (a
0)
{
map[i].w-=tmp;
map[i^1].w+=tmp;
cost+=tmp;
if(cost==lim)
break;
}
else
dis[v]=-1;
}
}
return cost;
}
int dinic()
{
int ans=0,s=0;
while(bfs())
ans+=dfs(s,inf);
//printf("%d\n",ans);
return ans;
}
int main()
{
int i,j,T,sum,t1,t2,cas=1;
int s[505],e[505],p[505];
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
t1=N;t2=0;
sum=0;
for(i=1;i<=n;i++)
{
scanf("%d%d%d",&p[i],&s[i],&e[i]);
t1=min(t1,s[i]);
t2=max(t2,e[i]);
sum+=p[i];
}
cnt=0;
memset(head,-1,sizeof(head));
for(i=t1;i<=t2;i++) //超级源点到一般源点的流量
{
add(0,i,m);
}
for(i=1;i<=n;i++)
{
for(j=s[i];j<=e[i];j++)
{
add(j,j+t2,1);
add(j+t2,2*t2,1);
}
}
t=t2*2;
if(sum<=dinic())
printf("Case %d: Yes\n\n",cas++);
else
printf("Case %d: No\n\n",cas++);
}
return 0;
}