HDU-2883 kebab (max Stream)

HDU-2883 kebab (max Stream)

There is a barbecue boss who can complete M's barbecue in every unit time
Now there are N guests who provide the time, time, quantity, and unit time required for each barbecue.
Ask the boss if he can fulfill the requirements of every guest.

Solution: This question is similar to HDU 3572, but it cannot be done like that, because it takes a long time.
Therefore, the time interval is treated as a point, and the range is connected to a super settlement point. The capacity is * M in length.
Connects all guests to a super origin site with a capacity of * time required for each barbecue
The next difficulty is how to connect the guests to the time range.
If the time range is between the time when the guests arrive and the time range, it indicates that the time range can be used to help the guests barbecue, so they can be connected, with the capacity of INF
In this way, the image is created.
Attach a detailed question
Detailed question

#include 
  
   #include 
   
    #include #include 
    
     #include 
     
      using namespace std;#define N 1010#define INF 0x3f3f3f3fstruct Edge {    int from, to, cap, flow;    Edge() {}    Edge(int from, int to, int cap, int flow): from(from), to(to), cap(cap), flow(flow) {}};struct ISAP {    int p[N], num[N], cur[N], d[N];    int t, s, n, m;    bool vis[N];    vector
      
        G[N]; vector
       
         edges; void init(int n) { this->n = n; for (int i = 0; i <= n; i++) { G[i].clear(); d[i] = INF; } edges.clear(); } void AddEdge(int from, int to, int cap) { edges.push_back(Edge(from, to, cap, 0)); edges.push_back(Edge(to, from, 0, 0)); m = edges.size(); G[from].push_back(m - 2); G[to].push_back(m - 1); } bool BFS() { memset(vis, 0, sizeof(vis)); queue
        
          Q; d[t] = 0; vis[t] = 1; Q.push(t); while (!Q.empty()) { int u = Q.front(); Q.pop(); for (int i = 0; i < G[u].size(); i++) { Edge &e = edges[G[u][i] ^ 1]; if (!vis[e.from] && e.cap > e.flow) { vis[e.from] = true; d[e.from] = d[u] + 1; Q.push(e.from); } } } return vis[s]; } int Augment() { int u = t, flow = INF; while (u != s) { Edge &e = edges[p[u]]; flow = min(flow, e.cap - e.flow); u = edges[p[u]].from; } u = t; while (u != s) { edges[p[u]].flow += flow; edges[p[u] ^ 1].flow -= flow; u = edges[p[u]].from; } return flow; } int Maxflow(int s, int t) { this->s = s; this->t = t; int flow = 0; BFS(); if (d[s] > n) return 0; memset(num, 0, sizeof(num)); memset(cur, 0, sizeof(cur)); for (int i = 0; i < n; i++) if (d[i] < INF) num[d[i]]++; int u = s; while (d[s] <= n) { if (u == t) { flow += Augment(); u = s; } bool ok = false; for (int i = cur[u]; i < G[u].size(); i++) { Edge &e = edges[G[u][i]]; if (e.cap > e.flow && d[u] == d[e.to] + 1) { ok = true; p[e.to] = G[u][i]; cur[u] = i; u = e.to; break; } } if (!ok) { int Min = n; for (int i = 0; i < G[u].size(); i++) { Edge &e = edges[G[u][i]]; if (e.cap > e.flow) Min = min(Min, d[e.to]); } if (--num[d[u]] == 0) break; num[d[u] = Min + 1]++; cur[u] = 0; if (u != s) u = edges[p[u]].from; } } return flow; }};ISAP isap;int S[N], E[N], num[N], T[N], All[N];int n, m;void solve() { int t, cnt = 0, s = 0, Sum = 0; for (int i = 1; i <= n; i++) { scanf(%d%d%d%d, &S[i], &num[i], &E[i], &T[i]); All[cnt++] = S[i]; All[cnt++] = E[i]; Sum += num[i] * T[i]; } sort(All, All + cnt); cnt = unique(All, All + cnt) - All; t = n + cnt + 1; isap.init(t); for (int i = 1; i <= n; i++) isap.AddEdge(s, i, num[i] * T[i]); for (int i = 1; i < cnt; i++) { isap.AddEdge(i + n, t, (All[i] - All[i - 1]) * m); for (int j = 1; j <= n; j++) { if (S[j] <= All[i - 1] && E[j] >= All[i]) { isap.AddEdge(j, i + n, INF); } } } if (Sum == isap.Maxflow(s, t)) printf(Yes); else printf(No);}int main() { while (scanf(%d%d, &n, &m) != EOF) { solve(); } return 0;}