HDU 3572 "Max Stream + array open small meeting tle"

Test instructions: There are M machines, there are n tasks. Each task must be done in SI or later, completed before EI or, the completion task must process the PI time unit. where each task can work on any (idle) machine, each machine can only work one task at a time, and each task can only be worked by one machine at a time, and the task is half-interrupted and taken to other machines. Can you finish the task within the stipulated time?

The map is simply classic. Just beginning to see is also a face, also want to use greedy to engage in a mess. But the vegetarian feel can't do it.
The correct idea is to use the maximum flow of the network stream.
S is connected with each task, the capacity is P, each task is connected with the corresponding time period, the capacity is 1, each point of time with the T edge, the capacity is M.
is not very clever, a look to understand, but completely can't think out

#include <stdio.h> #include <string.h> #include <queue> #include <iostream> #define MAX 510000# Define INF 0x3f3f3f3fusing namespace Std;int ss, Tt;int kase;int sum;int N, m;int cont;int head[max];int divv[max];int cur [max];struct Edge {int from, to, W, next;}    e[max];void Add (int u, int v, int w) {e[cont].from = u;    E[cont].to = v;    E[CONT].W = W;    E[cont].next = Head[u]; Head[u] = cont++;}    int Makediv () {memset (DIVV, 0, sizeof (DIVV));    DIVV[SS] = 1;    Queue<int> Q;    Q.push (ss); while (!        Q.empty ()) {int u = q.front ();        if (U = = TT) return 1;        Q.pop ();            for (int i = head[u]; i =-1; i = e[i].next) {int w = E[I].W;            int v = e[i].to;                if (divv[v] = = 0 && w) {divv[v] = Divv[u] + 1;            Q.push (v); }}} return 0;}    int DFS (int u, int maxflow, int tt) {if (U = = TT) return maxflow;    int ret = 0; for (int &Amp;i = Cur[u]; I! =-1;        i = e[i].next) {int v = e[i].to;        int w = E[I].W;            if (divv[v] = = Divv[u] + 1 && w) {int f = DFS (V, Min (Maxflow-ret, W), TT);            E[I].W-= f;            e[i ^ 1].W + = f;            RET + = f;        if (ret = = Maxflow) return ret;    }} if (ret) return ret;        else {Divv[u] = 0;    return 0;    }}void dinic () {int ans = 0;        while (makediv () = = 1) {memcpy (cur, head, sizeof (head));    Ans + = DFS (ss, INF, TT);    } if (ans = = sum) printf ("Case%d:yes\n\n", ++kase); else printf ("Case%d:no\n\n", ++kase);}    int main (void) {int t;    scanf ("%d", &t);    Kase = 0;        while (t--) {sum = 0;        SS = 0;        Cont = 0;        Memset (Head,-1, sizeof (head));        scanf ("%d%d", &n, &m);        int maxx = 0;        tt = 2 + n + +;        int p, S, E; for (int i = 1; I <= n; i++) {scanf ("%d%d%d ", &p, &s, &e);            sum + = p;            if (E > Maxx) maxx = e;            Add (ss, i + A, p);            Add (i + A, SS, 0);                for (int j = s; J <= E; j + +) {Add (i + A, J, 1);            Add (J, i + 500, 0);            }} for (int i = 1; I <= Maxx; i++) {Add (I, TT, M);        Add (TT, I, 0);    } dinic (); } return 0;}

HDU 3572 "Max Stream + array open small meeting tle"