Task Schedule

http://acm.hdu.edu.cn/showproblem.php?pid=3572

题意：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。

题解：网络流+最大流

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.

C++版本一

题解：Dinic算法+当前弧优化

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。

/*
*@Author:   STZG
*@Language: C++
*/

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
#include<ctime>
//#define DEBUG
#define RI register int
#define endl '\n'
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=1000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int s,t,n,m,k,p,l,r,u,v,w,c2;
int ans,cnt,flag,temp,sum;
int dis[N],cur[N];
struct node{
    int u,v,c;
    node(){};
    node(int form,int to,int cap):u(form),v(to),c(cap){}
};
vector<node>edge;
vector<int> G[N];
void Addedge(int u,int v,int cap){
    edge.push_back({u,v,cap});
    edge.push_back({v,u,0});
    int sz=edge.size();
    G[u].push_back(sz-2);
    G[v].push_back(sz-1);
}
bool bfs(int u){
    memset(dis,-1,sizeof(dis));
    dis[u]=0;
    queue<int>q;
    q.push(u);
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int i=0;i<G[u].size();i++){
            node e=edge[G[u][i]];//cout<<u<<" "<<e.v<<endl;
            if(dis[e.v]<0&&e.c>0){
                dis[e.v]=dis[u]+1;
                q.push(e.v);
            }
        }
    }
    return dis[t]>0;
}
int dfs(int u,int flow){
    if(u==t)
        return flow;
    int now;
    for(int &i=cur[u];i<G[u].size();i++){
        node e=edge[G[u][i]];
        if(e.c>0&&dis[u]+1==dis[e.v]&&(now=dfs(e.v,min(flow,e.c)))){
            edge[G[u][i]].c-=now;
            edge[G[u][i]^1].c+=now;
            return now;
        }
    }
    return 0;
}
void dinic(){
    while(bfs(s)){
        int res=0;
        memset(cur,0,sizeof(cur));
        while(res=dfs(s,INF)){
            ans+=res;
        }
    }
}
void init(){
    s=0;t=n+500+1;
    for(int i=s;i<=t;i++)G[i].clear();
    edge.clear();
    ans=0;
    sum=0;
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    scanf("%d",&k);
    int T=0;
    while(k--){
        scanf("%d%d",&n,&m);
        init();
        for(int i=1;i<=500;i++)Addedge(i+n,t,m);
        for(int i=1;i<=n;i++){
            scanf("%d%d%d",&p,&u,&v);
            Addedge(s,i,p);
            sum+=p;
            for(int j=u;j<=v;j++)
                Addedge(i,j+n,1);
        }
        dinic();
        //cout<<ans<<endl;
        cout<<"Case "<<++T<<": "<<(ans==sum?"Yes":"No")<<endl;
        cout<<endl;
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

C++版本二 

题解：ISAP算法

#include<iostream>
#include<cmath>
#include<cstring> 
using namespace std;   
#define MAXN 500 
#define MAXE 500002 
#define INF 0x7ffffff      
int ne,nv,tmp,s,t,index;  
struct Edge{ 
    int next,pair; 
    int v,cap,fLow; 
}edge[MAXE];
int net[MAXN];
int maxday; 
int ISAP() 
{ 
	int numb[MAXN],dist[MAXN],curedge[MAXN],pre[MAXN]; 
 
    int cur_fLow,max_fLow,u,tmp,neck,i; 
    memset(dist,0,sizeof(dist)); 
    memset(numb,0,sizeof(numb)); 
    memset(pre,-1,sizeof(pre)); 
    for(i = 1 ; i <= nv ; ++i) 
        curedge[i] = net[i]; 
    numb[nv] = nv; 
    max_fLow = 0; 
    u = s; 
    while(dist[s] < nv) 
    { 
        if(u == t) 
        { 
            cur_fLow = INF; 
            for(i = s; i != t;i = edge[curedge[i]].v)  
            {   
                if(cur_fLow > edge[curedge[i]].cap) 
                { 
                    neck = i; 
                    cur_fLow = edge[curedge[i]].cap; 
                } 
            } 
            for(i = s; i != t; i = edge[curedge[i]].v) 
            { 
                tmp = curedge[i]; 
                edge[tmp].cap -= cur_fLow; 
                edge[tmp].fLow += cur_fLow; 
                tmp = edge[tmp].pair; 
                edge[tmp].cap += cur_fLow; 
                edge[tmp].fLow -= cur_fLow; 
            } 
            max_fLow += cur_fLow; 
            u = neck; 
        } 
        /* if .... eLse ... */
        for(i = curedge[u]; i != -1; i = edge[i].next) 
            if(edge[i].cap > 0 && dist[u] == dist[edge[i].v]+1) 
                break; 
        if(i != -1) 
        { 
            curedge[u] = i; 
            pre[edge[i].v] = u; 
            u = edge[i].v; 
        }else{ 
            if(0 == --numb[dist[u]]) break; 
            curedge[u] = net[u]; 
            for(tmp = nv,i = net[u]; i != -1; i = edge[i].next) 
                if(edge[i].cap > 0) 
                    tmp = tmp<dist[edge[i].v]?tmp:dist[edge[i].v]; 
            dist[u] = tmp + 1; 
            ++numb[dist[u]]; 
            if(u != s) u = pre[u]; 
        } 
    }      
    return max_fLow; 
}
void addedge(int u,int v,int f)
{
	edge[index].next = net[u]; 
    edge[index].v = v; 
    edge[index].cap = f; 
    edge[index].fLow = 0; 
    edge[index].pair = index+1; 
    net[u] = index++; 
    edge[index].next = net[v]; 
    edge[index].v = u; 
    edge[index].cap = 0; 
    edge[index].fLow = 0; 
    edge[index].pair = index-1; 
    net[v] = index++;
	
} 
int main() 
{ 
    int i,j,np,nc,m,n; 
    int a,b,d,k,vaL;
	int tt;
	scanf("%d",&tt);
	for(int cas=1;cas<=tt;++cas)
	{ 
		int cases=1;
		int maxval=0;
		maxday=0;
		scanf("%d%d",&n,&m);  
		index=0;//index从0开始扫 
        s = 0; 
        t = 0;  
        memset(net,-1,sizeof(net));        
		for(i=1;i<=n;i++)
		{
			int pi,si,ei;
			scanf("%d%d%d",&pi,&si,&ei);
			maxday=max(maxday,ei);
			maxval+=pi;
			addedge(s,i,pi);
			for(j=si;j<=ei;j++)
				addedge(i,n+j,1);
			
			 
		}
		t=n+maxday+1;
		nv=t+1;
		for(i=1;i<=maxday;i++)
		addedge(i+n,t,m);		
		int ans=ISAP();
		if (ans == maxval)  
            printf("Case %d: Yes\n\n", cas);  
        else printf("Case %d: No\n\n", cas);  
 
    } 
    return 0; 
}