hdu 3572 Task Schedule（最大流，判断满流+isap模版）

Task Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2371    Accepted Submission(s): 858

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

24 31 3 5 1 1 42 3 73 5 92 22 1 31 2 2

 

Sample Output

Case 1: Yes   Case 2: Yes
题意：有n个任务和m台机器。每个任务都分别给出pi，si，ei，表示该任务只能在第si到ei天进行，且任务必须进行pi天才能完成。每个任务可以分成pi个部分，每个部分的进行不一定是连续的；每个任务都可以在不同的机器进行，一台机器同一时间只能做一个任务。问能否在约束条件下完成这n个任务。
思路：最大流判断是否满流。
把每个任务看成一个点，源点s到每个任务连边，容量为任务需要运行的天数。把每天看成一个点，每天向汇点t连边，容量为m，表示一天最多运行m个任务。每个任务都向s—e天连边，容量为1，表示这个任务可以在这些天运行。最后判断最大流是否等于所有任务需要运行的天数总和即可。
AC代码：
#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <stack>#include <vector>#include <cmath>#include <cstdlib>#define L(rt) (rt<<1)#define R(rt) (rt<<1|1)#define ll long long#define eps 1e-6using namespace std;const int INF=1000000000;const int maxn=1005;int eh[maxn],dis[maxn],low[maxn],cnt[maxn],pre[maxn],cur[maxn];int n,m,s,t,num;struct Edge{    int u,v,cap,flow,next;}et[maxn*maxn];void init(){    memset(eh,-1,sizeof(eh));    num=0;}void add(int u,int v,int cap,int flow){    Edge e={u,v,cap,flow,eh[u]};    et[num]=e;    eh[u]=num++;}void addedge(int u,int v,int cap){    add(u,v,cap,0);    add(v,u,0,0);}int isap(int s,int t,int n){    int u,v,now,flow=0;    memset(dis,0,sizeof(dis));    memset(cnt,0,sizeof(cnt));    memset(low,0,sizeof(low));    for(u=0;u<=n;u++) cur[u]=eh[u];    low[s]=INF,cnt[0]=n,u=s;    while(dis[s]<n)    {        for(now=cur[u];now!=-1;now=et[now].next)        if(et[now].cap-et[now].flow&&dis[u]==dis[v=et[now].v]+1) break;        if(now!=-1)        {            cur[u]=pre[v]=now;            low[v]=min(et[now].cap-et[now].flow,low[u]);            u=v;            if(u==t)            {                for(;u!=s;u=et[pre[u]].u)                {                    et[pre[u]].flow+=low[t];                    et[pre[u]^1].flow-=low[t];                }                flow+=low[t];                low[s]=INF;            }        }        else        {            if(--cnt[dis[u]]==0) break;            dis[u]=n,cur[u]=eh[u];            for(now=eh[u];now!=-1;now=et[now].next)            if(et[now].cap-et[now].flow&&dis[u]>dis[et[now].v]+1)            dis[u]=dis[et[now].v]+1;            cnt[dis[u]]++;            if(u!=s) u=et[pre[u]].u;        }    }    return flow;}int main(){    int ca,pi,si,ei,c=0;    scanf("%d",&ca);    while(ca--)    {        scanf("%d%d",&n,&m);        init();        s=0;        int sum=0,max_day=0;        for(int i=1; i<=n; i++)        {            scanf("%d%d%d",&pi,&si,&ei);            sum+=pi;            max_day=max(max_day,ei);            addedge(s,i,pi);            for(int j=si;j<=ei;j++) addedge(i,j+n,1);        }        t=max_day+1+n;        for(int i=1; i<=max_day; i++) addedge(i+n,t,m);        printf("Case %d: ",++c);        if(sum==isap(s,t,t+1)) printf("Yes\n");        else printf("No\n");        puts("");    }    return 0;}