hdu 3572 Task Schedule（当前弧优化Dinic算法）

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

解题思路：此题关键在于建图，跑最大流来判断是否已达到满流。题意：有n个任务，m台机器。每个任务有最早才能开始做的时间s_i，截止时间e_i，和完成该任务所需要的时间p_i。每个任务可以分段进行，但在同一天一台机器最多只能执行一个任务，问是否有可行的工作时间来完成所有任务。做法：建立一个超级源点s=0和一个超级汇点t=1001。源点s和每个任务i建边，边权为p_i，表示完成该任务需要的天数。对于每一个任务i，将编号i与编号n+[s_i～e_i]中每个编号建边，边权为1，表示将任务i分在第n+s_i天～第n+e_i天来完成，再将编号n+[s_i～e_i]与汇点t建边，边权为m，表示每一天(编号为n+j)最多同时运行m台机器来完成8天的任务单位量。但由于建的边数太多，用裸的Dinic跑最大流会TLE，怎么优化呢？采用当前弧优化：因为每次dfs找增广路的过程中都是从每个顶点指向的第1(编号为0)条边开始遍历的，而如果第1条边已达到满流，则会继续遍历第2条边....直至找到汇点，事实上这个递归的过程就造成了很多不必要浪费的时间，所以在dfs的过程中应标记一下每个顶点v当前能到达的第curfir[v]条边，说明顶点v的第0条边～第curfir[v]-1条边都已达满流或者流不到汇点，这样时间复杂度就大大降低了。注意：每次bfs给图重新分层次时，需要清空当前弧数组cirfir[]，表示初始时从每个顶点v的第1(编号为0)条边开始遍历。

AC代码(124ms)：

#include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=1005;
struct edge{ int to,cap;size_t rev;
    edge(int _to, int _cap, size_t _rev):to(_to),cap(_cap),rev(_rev){}
};
int T,n,m,p,s,e,tot,level[maxn];queue<int> que;vector<edge> G[maxn];size_t curfir[maxn];//当前弧数组
void add_edge(int from,int to,int cap){
    G[from].push_back(edge(to,cap,G[to].size()));
    G[to].push_back(edge(from,0,G[from].size()-1));
}
bool bfs(int s,int t){
    memset(level,-1,sizeof(level));
    while(!que.empty())que.pop();
    level[s]=0;
    que.push(s);
    while(!que.empty()){
        int v=que.front();que.pop();
        for(size_t i=0;i<G[v].size();++i){
            edge &e=G[v][i];
            if(e.cap>0&&level[e.to]<0){
                level[e.to]=level[v]+1;
                que.push(e.to);
            }
        }
    }
    return level[t]<0?false:true;
}
int dfs(int v,int t,int f){
    if(v==t)return f;
    for(size_t &i=curfir[v];i<G[v].size();++i){//从v的第curfir[v]条边开始，采用引用的方法，同时改变本身的值
        //因为节点v的第0~curfir[v]-1条边已达到满流了，所以无需重新遍历--->核心优化
        edge &e=G[v][i];
        if(e.cap>0&&(level[v]+1==level[e.to])){
            int d=dfs(e.to,t,min(f,e.cap));
            if(d>0){
                e.cap-=d;
                G[e.to][e.rev].cap+=d;
                return d;
            }
        }
    }
    return 0;
}
int max_flow(int s,int t){
    int f,flow=0;
    while(bfs(s,t)){
        memset(curfir,0,sizeof(curfir));//重新将图分层之后就清空数组，从第0条边开始遍历
        while((f=dfs(s,t,INF))>0)flow+=f;
    }
    return flow;
}
int main(){
    while(~scanf("%d",&T)){
        for(int cas=1;cas<=T;++cas){
            scanf("%d%d",&n,&m);tot=0;
            for(int i=0;i<maxn;++i)G[i].clear();
            for(int i=1;i<=500;++i)add_edge(500+i,1001,m);
            for(int i=1;i<=n;++i){
                scanf("%d%d%d",&p,&s,&e);
                add_edge(0,i,p);tot+=p;//tot为总时间
                for(int j=s;j<=e;++j)add_edge(i,500+j,1);
            }
            printf("Case %d: %s

",cas,max_flow(0,1001)==tot?"Yes":"No");
        }
    }
    return 0;
}