图论--网络流--最大流 HDU 3572 Task Schedule（限流建图，超级源汇）
图论--网络流--最大流 HDU 3572 Task Schedule（限流建图，超级源汇）
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
每个机器每台只能执行一个任务，每个任务在同一时段也只能被一台机执行。 给每个任务的开始时间和截止时间，和持续天数。最多给500天。
建立超级源点，源点到每个任务的流量为持续时间，每天到超级汇点的流量为M，这样能限制流量，即每天只能只能有M机器工作，然后每个任务到日期内的每一天设置流量为1，限制流量，即每天这个任务最多被一台机器干。欧克收工。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#define INF 1e9
using namespace std;
const int maxn =1000+10;
struct Edge
{
int from,to,cap,flow;
Edge(){}
Edge(int f,int t,int c,int fl):from(f),to(t),cap(c),flow(fl){}
};
struct Dinic
{
int n,m,s,t;
vector<Edge> edges;
vector<int> G[maxn];
bool vis[maxn];
int cur[maxn];
int d[maxn];
void init(int n,int s,int t)
{
this->n=n, this->s=s, this->t=t;
edges.clear();
for(int i=0;i<n;++i) G[i].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) );
m=edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
bool BFS()
{
queue<int> Q;
memset(vis,0,sizeof(vis));
vis[s]=true;
d[s]=0;
Q.push(s);
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]=true;
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[e.to]==d[x]+1 && (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 max_flow()
{
int ans=0;
while(BFS())
{
memset(cur,0,sizeof(cur));
ans +=DFS(s,INF);
}
return ans;
}
}DC;
int full_flow;
int main()
{
int T; scanf("%d",&T);
for(int kase=1;kase<=T;++kase)
{
int n,m;
scanf("%d%d",&n,&m);
full_flow=0;
int src=0,dst=500+n+1;
DC.init(500+2+n,src,dst);
bool vis[maxn];//表示第i天是否被用到
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;++i)
{
int P,S,E;
scanf("%d%d%d",&P,&S,&E);
DC.AddEdge(src,500+i,P);
full_flow += P;
for(int j=S;j<=E;++j)
{
DC.AddEdge(500+i,j,1);
vis[j]=true;
}
}
for(int i=1;i<=500;++i)if(vis[i])//被任务覆盖的日子才添加边
DC.AddEdge(i,dst,m);
printf("Case %d: %s\n\n",kase,DC.max_flow()==full_flow?"Yes":"No");
}
return 0;
}