[hdu3572]最大流(dinic)

题意：有m台机器，n个任务，每个任务需要在第si~ei天之间，且需要pi天才能完成，每台机器每天只能做一个任务，不同机器每天不能做相同任务，判断所有任务是否可以做完。

思路： 把影响答案的对象提取出来，得到以下几个：机器，任务，时间；需要用一个量把这三者联系起来，不难想到用工作量来表示。从源点向每个任务连一条容量为pi的有向边，表示这个任务需要pi个工作量才能完成，从每个任务向第si天到第ei天各连一条容量为1的有向边，表示这个任务可以在第si天到第ei天的任意一天“消耗”1个工作量，或者说第si天到第ei天的任意一天都可以花一个工作量来做这个工作，从每一天向汇点连一条容量为m的边，表示每一天允许产生m个工作量（m台机器每天产生m个工作量）。跑一遍最大流，看最大流是否等于所有任务的pi的和即可。

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/* ******************************************************************************** */
#include <iostream>                                                                 //
#include <cstdio>                                                                   //
#include <cmath>                                                                    //
#include <cstdlib>                                                                  //
#include <cstring>                                                                  //
#include <vector>                                                                   //
#include <ctime>                                                                    //
#include <deque>                                                                    //
#include <queue>                                                                    //
#include <algorithm>                                                                //
#include <map>                                                                      //
using namespace std;                                                                //
                                                                                    //
#define pb push_back                                                                //
#define mp make_pair                                                                //
#define X first                                                                     //
#define Y second                                                                    //
#define all(a) (a).begin(), (a).end()                                               //
#define foreach(a, i) for (typeof(a.begin()) i = a.begin(); i != a.end(); ++ i)     //
#define fill(a, x) memset(a, x, sizeof(a))                                          //
                                                                                    //
void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}    //
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>                    //
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;          //
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>      //
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>              //
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>   //
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}   //
                                                                                    //
typedef pair<int, int> pii;                                                         //
typedef long long ll;                                                               //
typedef unsigned long long ull;                                                     //
                                                                                    //
template<typename T>bool umax(T&a, const T&b){return b>a?false:(a=b,true);}         //
template<typename T>bool umin(T&a, const T&b){return b<a?false:(a=b,true);}         //
template<typename T>                                                                //
void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];}            //
template<typename T>                                                                //
void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];}            //
                                                                                    //
/* -------------------------------------------------------------------------------- */
 
 
struct Dinic {
private:
    const static int maxn = 1e3 + 7;
    struct Edge {
        int from, to, cap;
        Edge(int u, int v, int w): from(u), to(v), cap(w) {}
    };
    int s, t;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool vis[maxn];
    int d[maxn], cur[maxn];
 
    bool bfs() {
        memset(vis, 0, sizeof(vis));
        queue<int> Q;
        Q.push(s);
        d[s] = 0;
        vis[s] = true;
        while (!Q.empty()) {
            int x = Q.front(); Q.pop();
            for (int i = 0; i < G[x].size(); i ++) {
                Edge &e = edges[G[x][i]];
                if (!vis[e.to] && e.cap) {
                    vis[e.to] = true;
                    d[e.to] = d[x] + 1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }
    int dfs(int x, int a) {
        if (x == t || a == 0) return a;
        int flow = 0, f;
        for (int &i = cur[x]; i < G[x].size(); i ++) {
            Edge &e = edges[G[x][i]];
            if (d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap))) > 0) {
                e.cap -= f;
                edges[G[x][i] ^ 1].cap += f;
                flow += f;
                a -= f;
                if (a == 0) break;
            }
        }
        return flow;
    }
 
public:
    void clear() {
        for (int i = 0; i < maxn; i ++) G[i].clear();
        edges.clear();
        memset(d, 0, sizeof(d));
    }
    void add(int from, int to, int cap) {
        edges.push_back(Edge(from, to, cap));
        edges.push_back(Edge(to, from, 0));
        int m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
    }
 
    int solve(int s, int t) {
        this->s = s; this->t = t;
        int flow = 0;
        while (bfs()) {
            memset(cur, 0, sizeof(cur));
            flow += dfs(s, 1e9);
        }
        return flow;
    }
};
Dinic solver;
const int maxn = 507;
int p[maxn], s[maxn], e[maxn];
 
int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif // ONLINE_JUDGE
    int T, n, m, cas = 0;
    cin >> T;
    while (T --) {
        cin >> n >> m;
        solver.clear();
        int total = 0;
        for (int i = 1; i <= n; i ++) {
            scanf("%d%d%d", p + i, s + i, e + i);
            total += p[i];
        }
        for (int i = 1; i <= n; i ++) {
            solver.add(0, i, p[i]);
            for (int j = s[i]; j <= e[i]; j ++) {
                solver.add(i, n + j, 1);
            }
        }
        for (int i = 1; i <= 500; i ++) solver.add(n + i, n + 501, m);
        printf("Case %d: ", ++ cas);
        puts(solver.solve(0, n + 501) == total? "Yes" : "No");
        puts("");
    }
    return 0;                                                                       //
}                                                                                   //
                                                                                    //
                                                                                    //
                                                                                    //
/* ******************************************************************************** */

 

posted @ 2015-07-30 07:07  jklongint  阅读(270)  评论(0)  编辑  收藏  举报