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
交了10遍,一直TLE,然后改成了邻接表建边dinic,然后又进行了优化,A了。。。
这个题意不难,将源点和任务相连,容量是需要的天数,然后将任务需要在允许的天数内建边,容量为1,最后将所有的天数和汇点建边,容量为m,因为每天最多有m个,因为每个任务至少一天,如果同时工作,所以最多为m,但是最重要的是明白dinic邻接表建边
#include<cstdio>
#include<cstring>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
#define oo 1<<28
using namespace std;
int n,m,start,end,cnt;
int dep[1100];
int p[1100];
struct node
{
int u,v,w;
int next;
} edge[1100000];
void add(int u,int v,int w)
{
edge[cnt].u=u;
edge[cnt].v=v;
edge[cnt].w=w;
edge[cnt].next=p[u];
p[u]=cnt++;
edge[cnt].u=v;
edge[cnt].v=u;
edge[cnt].w=0;
edge[cnt].next=p[v];
p[v]=cnt++;
}
int BFS()
{
queue<int>q;
memset(dep,-1,sizeof(dep));
dep[start]=0;
while(!q.empty())
{
q.pop();
}
q.push(start);
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=p[u]; i!=-1; i=edge[i].next)
{
if(edge[i].w>0 && dep[edge[i].v]==-1)// 如果可以到达且还没有访问
{
dep[edge[i].v]=dep[u]+1;
q.push(edge[i].v);
}
}
}
if(dep[end]>0)
return 1;
return 0;
}
int find(int i,int sum) // 查找路径上的最小的流量
{
if(i==end)
return sum;
int tmp;
int t=0;
for(int j=p[i]; j!=-1; j=edge[j].next)
{
int v=edge[j].v;
int w=edge[j].w;
if(w>0 && dep[v]==dep[i]+1 && (tmp=find(v,min(sum,w))))
{
edge[j].w-=tmp; //正向减少
edge[j^1].w+=tmp; //反向增加
t+=tmp;
sum-=tmp;
if(!sum)
break;
}
}
if(t)
return t;
dep[i]=-1;
return 0;
}
int Dinic()
{
int ans=0,tmp;
while(BFS())//如果能找到一条增广路
{
tmp=find(start,oo);//先把开始给标号,oo是因为没有节点控制,先初始为最
if(tmp==0)
break;
ans+=tmp;
}
return ans;
}
int main()
{
int t,i,j,k;
scanf("%d",&t);
for(k=1; k<=t; k++)
{
cnt=0;
scanf("%d%d",&n,&m);
memset(p,-1,sizeof(p));
int d,u,v,w;
int sum=0,maxx=0,minn=oo;
for(i=1; i<=n; i++)
{
scanf("%d%d%d",&d,&u,&v);
sum+=d;
if(minn>u)
minn=u;
if(maxx<v)
maxx=v;
add(0,i,d);//将源点和每一个任务相连,权值为每个任务需要的天数
for(j=u; j<=v; j++)
{
add(i,j+n,1);//将每一个任务与他们需要完成得天数区间内相连,权值为1,代表可在此天
}
}
int dd=n+maxx+1;//汇点
for(i=minn+n; i<=maxx+n; i++)
{
add(i,dd,m);//最后将每个天数与汇点相连,权值为m,表示这一天可以有运行m个任务,因为每个任务至少需要一个机器1天,故最多运行m个任务
}
start=0, end=dd;//开始于0,结束于dd
if(sum==Dinic())
printf("Case %d: Yes\n\n",k);
else
printf("Case %d: No\n\n",k);
}
return 0;
}