题意:给N个任务,M台机器。每个任务有起始时间S,最后期限 E,和完成这个任务需要的时间P。每个任务可以分段进行,但是在同一时刻,一台机器最多只能执行一个任务. 问存不存在可行的工作时间。
思路:这题由于每一个任务都是可以断断续续这样完成的,所以一开始不知道怎么建图,留意到这里一共的时间才500,所以可以把每一天都作为一个顶点,任务的编号为500+1到500+n,汇点为500+n+1
建图:
从源点0连一条容量为m的边到每一个点,表示每天最多可以有多少个机器可以工作,每个任务连向汇点,容量为完成任务需要的时间,枚举500天,如果该天满足大于等于起始时间,小于等于最后期限,就在这天和机器连一条边,容量为1,最后跑一遍最大流看看是否等于总时间即可
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 1550
#define INF 1<<29
#define LL long long
int cas=1,T;
struct Edge
{
int from,to,cap,flow;
Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}
};
int n,m;
struct Dinic
{
// int n,m;
int s,t;
vector<Edge>edges; //边数的两倍
vector<int> G[maxn]; //邻接表,G[i][j]表示结点i的第j条边在e数组中的序号
bool vis[maxn]; //BFS使用
int d[maxn]; //从起点到i的距离
int cur[maxn]; //当前弧下标
void init()
{
for (int i=0;i<=1200;i++)
G[i].clear();
edges.clear();
}
void AddEdge(int from,int to,int cap)
{
edges.push_back(Edge(from,to,cap,0));
edges.push_back(Edge(to,from,0,0)); //反向弧
int mm=edges.size();
G[from].push_back(mm-2);
G[to].push_back(mm-1);
}
bool BFS()
{
memset(vis,0,sizeof(vis));
queue<int>q;
q.push(s);
d[s]=0;
vis[s]=1;
while (!q.empty())
{
int x = q.front();q.pop();
for (int i = 0;i<G[x].size();i++)
{
Edge &e = edges[G[x][i]];
if (!vis[e.to] && e.cap > e.flow)
{
vis[e.to]=1;
d[e.to] = d[x]+1;
q.push(e.to);
}
}
}
return vis[t];
}
int DFS(int x,int a)
{
if (x==t || a==0)
return a;
int flow = 0,f;
for(int &i=cur[x];i<G[x].size();i++)
{
Edge &e = edges[G[x][i]];
if (d[x]+1 == d[e.to] && (f=DFS(e.to,min(a,e.cap-e.flow)))>0)
{
e.flow+=f;
edges[G[x][i]^1].flow-=f;
flow+=f;
a-=f;
if (a==0)
break;
}
}
return flow;
}
int Maxflow(int s,int t)
{
this->s=s;
this->t=t;
int flow = 0;
while (BFS())
{
memset(cur,0,sizeof(cur));
flow+=DFS(s,INF);
}
return flow;
}
}dc;
int p[maxn];
int s[maxn];
int e[maxn];
int main()
{
scanf("%d",&T);
while (T--)
{
printf("Case %d: ",cas++);
scanf("%d%d",&n,&m);
dc.init();
memset(p,0,sizeof(p));
memset(s,0,sizeof(s));
memset(e,0,sizeof(e));
int sum =0;
for (int i = 1;i<=n;i++)
{
scanf("%d%d%d",&p[i],&s[i],&e[i]);
sum+=p[i];
}
for (int i = 1;i<=500;i++) //源点到每一天连边
dc.AddEdge(0,i,m);
for (int i = 1;i<=n;i++)
dc.AddEdge(500+i,500+n+1,p[i]); //每个任务到汇点有边
for (int i = 1;i<=n;i++)
{
for (int j=1;j<=500;j++)
{
if (j <= e[i] && j>=s[i])
dc.AddEdge(j,500+i,1);
}
}
printf("%s\n\n",dc.Maxflow(0,500+n+1) == sum?"Yes":"No");
}
}
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