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.
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.
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.
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
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<b?a:b)
const int inf=0x7ffffff;
int cnt,n,m,t;
int head[N],q[N],dis[N];
struct node
{
int u,v,w,next;
}map[N*N];
void add(int u,int v,int w)
{
map[cnt].u=u;
map[cnt].v=v;
map[cnt].w=w;
map[cnt].next=head[u];
head[u]=cnt++;
map[cnt].u=v;
map[cnt].v=u;
map[cnt].w=0;
map[cnt].next=head[v];
head[v]=cnt++;
}
int bfs()
{
int i,u,v,t1,t2;
memset(dis,0,sizeof(dis));
u=t1=t2=0;
dis[u]=1;
q[t1++]=u;
while(t2<t1)
{
u=q[t2++];
for(i=head[u];i!=-1;i=map[i].next)
{
v=map[i].v;
if(map[i].w&&!dis[v])
{
dis[v]=dis[u]+1;
if(v==t)
return 1;
q[t1++]=v;
}
}
}
return 0;
}
int dfs(int s,int lim)
{
int i,tmp,v,cost=0;
if(s==t)
return lim;
for(i=head[s];i!=-1;i=map[i].next)
{
v=map[i].v;
if(map[i].w&&dis[s]==dis[v]-1)
{
tmp=dfs(v,min(lim-cost,map[i].w));
if(tmp>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;
}
版权声明：本文博客原创文章，博客，未经同意，不得转载。
本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/4720126.html，如需转载请自行联系原作者