HDU 2883 kebab

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 2000;
 4 const int INF = 0x3f3f3f3f;
 5 struct arc{
 6     int to,flow,next;
 7     arc(int x = 0,int y = 0,int z = -1){
 8         to = x;
 9         flow = y;
10         next = z;
11     }
12 }e[1000000];
13 struct Server{
14     int s,n,e,t;
15 }SV[500];
16 int d[maxn],head[maxn],p[maxn],cur[maxn],tot,S,T,n,m;
17 void add(int u,int v,int flow){
18     e[tot] = arc(v,flow,head[u]);
19     head[u] = tot++;
20     e[tot] = arc(u,0,head[v]);
21     head[v] = tot++;
22 }
23 bool bfs(){
24     queue<int>q;
25     memset(d,-1,sizeof d);
26     d[S] = 1;
27     q.push(S);
28     while(!q.empty()){
29         int u = q.front();
30         q.pop();
31         for(int i = head[u]; ~i; i = e[i].next){
32             if(e[i].flow && d[e[i].to] == -1){
33                 d[e[i].to] = d[u] + 1;
34                 q.push(e[i].to);
35             }
36         }
37     }
38     return d[T] > -1;
39 }
40 int dfs(int u,int low){
41     if(u == T) return low;
42     int tmp = 0,a;
43     for(int &i = cur[u]; ~i; i = e[i].next){
44         if(e[i].flow && d[e[i].to] == d[u]+1 &&(a=dfs(e[i].to,min(low,e[i].flow)))){
45             tmp += a;
46             low -= a;
47             e[i].flow -= a;
48             e[i^1].flow += a;
49             if(!low) break;
50         }
51     }
52     if(!tmp) d[u] = -1;
53     return tmp;
54 }
55 int dinic(){
56     int ret = 0;
57     while(bfs()){
58         memcpy(cur,head,sizeof head);
59         ret += dfs(S,INF);
60     }
61     return ret;
62 }
63 int main(){
64     while(~scanf("%d%d",&n,&m)){
65         memset(head,-1,sizeof head);
66         int cnt = 0;
67         for(int i = 1; i <= n; ++i){
68             scanf("%d%d%d%d",&SV[i].s,&SV[i].n,&SV[i].e,&SV[i].t);
69             p[cnt++] = SV[i].s;
70             p[cnt++] = SV[i].e;
71         }
72         sort(p,p+cnt);
73         cnt = unique(p,p+cnt)-p;
74         int sum = S = tot = 0;
75         T = cnt+n+1;
76         for(int i = 1; i < cnt; ++i) add(i+n,T,m*(p[i] - p[i-1]));
77         for(int i = 1; i <= n; ++i){
78             add(S,i,SV[i].n*SV[i].t);
79             sum += SV[i].n*SV[i].t;
80             for(int j = 1; j < cnt; ++j)
81                 if(SV[i].s <= p[j-1] && SV[i].e >= p[j]) add(i,j+n,INF);
82         }
83         puts(dinic() == sum?"Yes":"No");
84     }
85     return 0;
86 }