码迷,mamicode.com
首页 > 其他好文 > 详细

hdu 3572 Task Schedule(最大流)

时间:2015-08-02 21:40:22      阅读:103      评论:0      收藏:0      [点我收藏+]

标签:

hdu 3572 Task Schedule

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

题目大意:给出任务数量,和机器数量。每个任务包括三个信息:一台机器需要处理的时间,起始时间,终止时间。问能否在每个任务的时限内完成所有人物。

解题思路:完全没思路,觉得这应该是贪心的题目。后来去看了题解……。设置一个超级源点,连向所有的任务,容量为该任务需要处理的时间。然后每个任务连向他们对应的时间,例如一个任务时限从2到4,则该任务结点,连向2 3 4的时间节点,容量为1。然后所有的时间节点,连向超级汇点,容量为机器数量。挺巧妙的,但我想不到……

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 2500;
const int M = 400005;
const int FIN = 2015;
typedef long long ll;
int n, m, s, t, sum;

int ec, head[N], first[N], que[N], lev[N];
int Next[M], to[M], v[M];

void init() {
    sum = 0;
    ec = 0;
    memset(first, -1, sizeof(first));
    s = 0, t = FIN;
}

void addEdge(int a,int b,int c) {
    to[ec] = b;
    v[ec] = c;
    Next[ec] = first[a];
    first[a] = ec++;

    to[ec] = a;
    v[ec] = 0;
    Next[ec] = first[b];
    first[b] = ec++;
}

int BFS() {
    int kid, now, f = 0, r = 1, i;
    memset(lev, 0, sizeof(lev));
    que[0] = s, lev[s] = 1;
    while (f < r) {
        now = que[f++];
        for (i = first[now]; i != -1; i = Next[i]) {
            kid = to[i];    
            if (!lev[kid] && v[i]) {
                lev[kid] = lev[now] + 1;    
                if (kid == t) return 1;
                que[r++] = kid;
            }
        }
    }
    return 0;
}

int DFS(int now, int sum) {
    int kid, flow, rt = 0;
    if (now == t) return sum;
    for (int i = head[now]; i != -1 && rt < sum; i = Next[i]) {
        head[now] = i;  
        kid = to[i];
        if (lev[kid] == lev[now] + 1 && v[i]) {
            flow = DFS(kid, min(sum - rt, v[i]));
            if (flow) {
                v[i] -= flow;
                v[i^1] += flow;
                rt += flow;
            } else lev[kid] = -1;   
        }           
    }
    return rt;
}

int dinic() {
    int ans = 0;
    while (BFS()) {
        for (int i = 0; i <= t; i++) {
            head[i] = first[i];
        }           
        ans += DFS(s, INF);
    }
    return ans;
}   

void input() {
    int Min = INF, Max = 0;
    scanf("%d %d", &n, &m);
    int a, b, c;
    for (int i = 1; i <= n; i++) {
        scanf("%d %d %d", &a, &b, &c);  
        sum += a;
        addEdge(s, i, a);
        for (int j = b; j <= c; j++) {
            addEdge(i, j + n, 1);   
        }
        if (c < b) swap(c, b);
        if (Min > b) Min = b;
        if (Max < c) Max = c;
    }
    for (int i = Min; i <= Max; i++) {
        addEdge(i + n, t, m);   
    }
}

int main() {
    int T, Case = 1;
    scanf("%d", &T);
    while (T--) {
        printf("Case %d: ", Case++);
        init();
        input();
        if (dinic() == sum) printf("Yes\n");
        else printf("No\n");
        puts("");
    }
    return 0;
}

版权声明:本文为博主原创文章,未经博主允许也可以转载,不过要注明出处哦。

hdu 3572 Task Schedule(最大流)

标签:

原文地址:http://blog.csdn.net/llx523113241/article/details/47211211

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!