hdu 3572 Task Schedule (Dinic模板)
2024-09-05 22:41:36
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
给N个任务,M台机器。每个任务有最早才能开始做的时间S,deadline E,和持续工作的时间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。
由于时间<=500且每个任务都能断断续续的执行,那么我们把每一天时间作为一个节点来用网络流解决该题.
建图: 源点s(编号0), 时间1-500天编号为1到500, N个任务编号为500+1 到500+N, 汇点t(编号501+N).
源点s到每个任务i有边(s, i, Pi)
每一天到汇点有边(j, t, M) (其实这里的每一天不一定真要从1到500,只需要取那些被每个任务覆盖的每一天即可)
如果任务i能在第j天进行,那么有边(i, j, 1) 注意由于一个任务在一天最多只有1台机器执行,所以该边容量为1,不能为INF或M哦.
最后看最大流是否 == 所有任务所需要的总天数.
#include <bits/stdc++.h> using namespace std;
const int maxn = ;
const int inf = 0x3f3f3f3f;
struct Edge
{
int from,to,cap,flow;
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> edge;
vector <int> G[maxn];//存图
bool vis[maxn];//标记每点是否vis过
int cur[maxn];//当前弧优化
int dep[maxn];//标记深度
void init(int n,int s,int t)//初始化
{
this->n=n;this->s=s;this->t=t;
edge.clear();
for (int i=;i<n;++i) G[i].clear();
}
void addedge (int from,int to,int cap)//加边,单向边
{
edge.push_back(Edge(from,to,cap,));
edge.push_back(Edge(to,from,,));
m=edge.size();
G[from].push_back(m-);
G[to].push_back(m-);
}
bool bfs ()
{
queue<int> q;
while (!q.empty()) q.pop();
memset(vis,false,sizeof vis);
vis[s]=true;
dep[s]=;
q.push(s);
while (!q.empty()){
int u=q.front();
//printf("%d\n",u);
q.pop();
for (int i=;i<G[u].size();++i){
Edge e=edge[G[u][i]];
int v=e.to;
if (!vis[v]&&e.cap>e.flow){
vis[v]=true;
dep[v]=dep[u]+;
q.push(v);
}
}
}
return vis[t];
}
int dfs (int x,int mi)
{
if (x==t||mi==) return mi;
int flow=,f;
for (int &i=cur[x];i<G[x].size();++i){
Edge &e=edge[G[x][i]];
int y=e.to;
if (dep[y]==dep[x]+&&(f=dfs(y,min(mi,e.cap-e.flow)))>){
e.flow+=f;
edge[G[x][i]^].flow-=f;
flow+=f;
mi-=f;
if (mi==) break;
}
}
return flow;
}
int max_flow ()
{
int ans = ;
while (bfs()){
memset(cur,,sizeof cur);
ans+=dfs(s,inf);
}
return ans;
}
}dinic;
int full_flow;
int main()
{
int casee = ;
//freopen("de.txt","r",stdin);
int T;scanf("%d",&T);
while (T--){
int n,m;
full_flow = ;
scanf("%d%d",&n,&m);
int src = ,dst = ++n;
dinic.init(++n,src,dst);
bool vis[maxn];
memset(vis,false,sizeof vis);
for (int i=;i<=n;++i){
int p,s,e;
scanf("%d%d%d",&p,&s,&e);
full_flow+=p;
dinic.addedge(src,i+,p);
for (int j=s;j<=e;++j){
vis[j]=true;
dinic.addedge(i+,j,);
}
}
for (int i=;i<maxn;++i){
if (vis[i])
dinic.addedge(i,dst,m);
}
printf("Case %d: ",++casee);
if (dinic.max_flow()==full_flow){//dinic.max_flow()只能跑一遍
printf("Yes\n\n");
}
else
printf("No\n\n");
}
return ;
}
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