【最大流】 HDU 3572 Task Schedule

题意 给出N个任务和M台机器

任务中给出 C  S E 分别表示 所需时间  任务开始时间  任务结束时间 

即任务需要在开始到结束这时间段中完成。

一个任务可以在不同的机器上完成,但每次在机器上至少要运行一天

建边, 源点-任务 容量为C

   任务-时间段上的每一个时间点 容量为1 

   时间点-汇点 容量为 M

最大流 >= sigma(C)则YES else NO

(最大所需点数为n+m+2;)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <string>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <math.h>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
#include <time.h>;
#define cler(arr, val)    memset(arr, val, sizeof(arr))
#define IN     freopen ("in.txt" , "r" , stdin);
#define OUT  freopen ("out.txt" , "w" , stdout);
typedef long long  LL;
const int MAXN = 200010;//点数的最大值
const int MAXM = 200006;//边数的最大值
const int INF = 0x3f3f3f3f;
const int mod = 10000007;
struct Edge
{
    int to,next,cap,flow;
}edge[MAXM];//注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
void init()
{
    tol = 0;
    memset(head,-1,sizeof (head));
}
void addedge (int u,int v,int w,int rw = 0)
{
    edge[tol].to = v; edge[tol].cap = w; edge[tol].flow = 0;
    edge[tol].next = head[u]; head[u] = tol++;
    edge[tol].to = u; edge[tol].cap = rw; edge[tol].flow = 0;
    edge[tol].next = head[v]; head[v] = tol++;
}
int Q[MAXN];
void BFS(int start,int end)
{
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0] = 1;
    int front = 0, rear = 0;
    dep[end] = 0;
    Q[rear++] = end;
    while(front != rear)
    {
        int u = Q[front++];
        for(int i = head[u]; i !=  -1; i = edge[i].next)
        {
            int v = edge[i]. to;
            if(dep[v] != -1)continue;
            Q[rear++] = v;
            dep[v] = dep[u] + 1;
            gap[dep[v]]++;
        }
    }
}
int S[MAXN];
int sap(int start,int end, int N)
{
    BFS(start,end);
    memcpy(cur,head,sizeof(head));
    int top = 0;
    int u = start;
    int ans = 0;
    int i;
    while(dep[start] < N)
    {
        if(u == end)
        {
            int Min = INF;
            int inser;
            for( i = 0;i < top;i++)
            {
                if(Min > edge[S[i]].cap - edge[S[i]].flow)
                {
                    Min = edge[S[i]].cap - edge[S[i]].flow;
                    inser = i;
                }
            }
            for( i = 0;i < top;i++)
            {
                edge[S[i]]. flow += Min;
                edge[S[i]^1].flow -= Min;
            }
            ans += Min;
            top = inser;
            u = edge[S[top]^1].to;
            continue;
        }
        bool flag =  false;
        int v;
        for( i = cur[u]; i != -1; i = edge[i]. next)
        {
            v = edge[i]. to;
            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
            {
                flag =  true;
                cur[u] = i;
                break;
            }
        }
        if(flag)
        {
            S[top++] = cur[u];
            u = v;
            continue;
        }
        int Min = N;
        for( i = head[u]; i !=  -1; i = edge[i].next)
        {
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        }
        gap[dep[u]]--;
        if(!gap[dep[u]]) return ans;
        dep[u] = Min + 1;
        gap[dep[u]]++;
        if(u != start)u = edge[S[--top]^1].to;
    }
    return ans;
}
int main()
{
    //IN;
    int t,cas=0,m,n;
    scanf("%d",&t);
    while(t--)
    {
        init();
        scanf("%d %d",&n,&m);
        int e=500+n+1,sum=0;
        for(int i=1;i<=500;i++)
        {
            addedge(n+i,e,m);
        }
        for(int i=1;i<=n;i++)
        {
            int x,y,c;
            scanf("%d %d %d",&c,&x,&y);
            sum+=c;
            addedge(0,i,c);//源点到任务
            for(int j=x;j<=y;j++)//任务到天
            {
                addedge(i,n+j,1);
            }
        }
        printf("Case %d: ",++cas);
        int out=sap(0,e,n+500+2);
      //  printf("%d\n",out);
        if(out>=sum)
            puts("Yes");
        else puts("No");
        puts("");
    }
}


posted @ 2014-09-26 20:20  kewowlo  阅读(128)  评论(0编辑  收藏  举报