HDU3572Task Schedule(最大流 ISAP比較快)建图方法不错

Task Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5007    Accepted Submission(s): 1636

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

 

Author

allenlowesy

 

Source

2010 ACM-ICPC Multi-University Training Contest（13）——Host by UESTC

 

题意：有M个机器，有N个任务。

每一个任务必须在Si 或者以后開始做，在Ei 或者之前完毕。完毕任务必须处理Pi 个时间单位。当中，每一个任务能够在随意（空暇）机器上工作，每一个机器的同一时刻仅仅能工作一个任务。每一个任务在同一时刻仅仅能被一个机器工作，并且任务做到一半能够打断，拿去其它机器做。问：是否能在规定时间内把任务做完。

思路：建图是关键，我们能够选择0为源点，然后源点与每一个任务都连一条边。容量为要求的天数p,然后每一个任务都与对应的时间点连边。边容量为1。最后我们要确定汇点。汇点能够取vt=maxtime+n+1(当中maxtime为结束时间的最大值，n为任务数），这样确定好汇点之后，再在每一个时间点与汇点之间连边。边容量为m,为机器数(表示每一个时间点最多能够有m台机器处理任务)，最后推断sum(for all pi)==?SAP就可以。

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define captype int

const int MAXN = 1010;   //点的总数
const int MAXM = 520010;    //边的总数
const int INF = 1<<30;
struct EDG{
    int to,next;
    captype cap,flow;
} edg[MAXM];
int eid,head[MAXN];
int gap[MAXN];  //每种距离(或可觉得是高度)点的个数
int dis[MAXN];  //每一个点到终点eNode 的最短距离
int cur[MAXN];  //cur[u] 表示从u点出发可流经 cur[u] 号边

void init(){
    eid=0;
    memset(head,-1,sizeof(head));
}
//有向边 三个參数，无向边4个參数
void addEdg(int u,int v,captype c,captype rc=0){
    edg[eid].to=v; edg[eid].next=head[u];
    edg[eid].cap=c; edg[eid].flow=0; head[u]=eid++;

    edg[eid].to=u; edg[eid].next=head[v];
    edg[eid].cap=rc; edg[eid].flow=0; head[v]=eid++;
}
//预处理eNode点到全部点的最短距离
void BFS(int sNode, int eNode){
    queue<int>q;
    memset(gap,0,sizeof(gap));
    memset(dis,-1,sizeof(dis));
    gap[0]=1;
    dis[eNode]=0;
    q.push(eNode);
    while(!q.empty()){
        int u=q.front(); q.pop();
        for(int i=head[u]; i!=-1; i=edg[i].next){
            int v=edg[i].to;
            if(dis[v]==-1){
                dis[v]=dis[u]+1;
                gap[dis[v]]++;
                q.push(v);
            }
        }
    }
}
int S[MAXN];    //路径栈。存的是边的id号
captype maxFlow_sap(int sNode,int eNode, int n){
    BFS(sNode, eNode);              //预处理eNode到全部点的最短距离
    if(dis[sNode]==-1) return 0;    //源点到不可到达汇点
    memcpy(cur,head,sizeof(head));

    int top=0;  //栈顶
    captype ans=0;  //最大流
    int u=sNode;
    while(dis[sNode]<n){   //推断从sNode点有没有流向下一个相邻的点
        if(u==eNode){   //找到一条可增流的路
            captype Min=INF ;
            int inser;
            for(int i=0; i<top; i++)    //从这条可增流的路找到最多可增的流量Min
            if(Min>edg[S[i]].cap-edg[S[i]].flow){
                Min=edg[S[i]].cap-edg[S[i]].flow;
                inser=i;
            }
            for(int i=0; i<top; i++){
                edg[S[i]].flow+=Min;
                edg[S[i]^1].flow-=Min;  //可回流的边的流量
            }
            ans+=Min;
            top=inser;  //从这条可增流的路中的流量瓶颈 边的上一条边那里是能够再增流的。所以仅仅从断流量瓶颈 边裁断
            u=edg[S[top]^1].to;  //流量瓶颈 边的起始点
            continue;
        }
        bool flag = false;  //推断是否能从u点出发可往相邻点流
        int v;
        for(int i=cur[u]; i!=-1; i=edg[i].next){
            v=edg[i].to;
            if(edg[i].cap-edg[i].flow>0 && dis[u]==dis[v]+1){
                flag=true;
                cur[u]=i;
                break;
            }
        }
        if(flag){
            S[top++] = cur[u];  //增加一条边
            u=v;
            continue;
        }
        //假设上面没有找到一个可流的相邻点。则改变出发点u的距离（也可觉得是高度）为相邻可流点的最小距离+1
        int Mind= n;
        for(int i=head[u]; i!=-1; i=edg[i].next)
        if(edg[i].cap-edg[i].flow>0 && Mind>dis[edg[i].to]){
            Mind=dis[edg[i].to];
            cur[u]=i;
        }
        gap[dis[u]]--;
        if(gap[dis[u]]==0) return ans;  //当dis[u]这样的距离的点没有了，也就不可能从源点出发找到一条增广流路径
                                        //由于汇点到当前点的距离仅仅有一种，那么从源点到汇点必定经过当前点，然而当前点又没能找到可流向的点，那么必定断流
        dis[u]=Mind+1;      //假设找到一个可流的相邻点。则距离为相邻点距离+1，假设找不到，则为n+1
        gap[dis[u]]++;
        if(u!=sNode) u=edg[S[--top]^1].to;  //退一条边

    }
    return ans;
}
int main(){
    int T,cas=0,n,m;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        init();
        int maxtime=0,sump=0;
        for(int i = 1; i<=n; i++){
            int P,S,E;
            scanf("%d%d%d",&P,&S,&E);

            sump += P;   //总流量
            if(E>maxtime)
                maxtime = E;

            addEdg(0,i,P);  //源点0到任务i的最大流量为P
            for(int j=S; j<=E; j++)
                addEdg(i,n+j,1);   //任务i到时间点j+n。边最大容量为1
        }
        int t=maxtime + n +1;
        for(int j=1; j<=maxtime; j++)
            addEdg(j+n,t,m);<span style="white-space:pre">	</span>//每一个时间点到汇点t，边最大容量为m

        printf("Case %d: %s\n\n",++cas,maxFlow_sap(0,t,t)==sump?

"Yes":"No"); } }