题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2883
Almost everyone likes kebabs nowadays (Here a kebab means pieces of meat grilled on a long thin stick). Have you, however, considered about the hardship of a kebab roaster while enjoying the delicious food? Well, here's a chance for you to help the poor roaster make sure whether he can deal with the following orders without dissatisfying the customers.
Now N customers is coming. Customer i will arrive at time si (which means the roaster cannot serve customer i until time si). He/She will order ni kebabs, each one of which requires a total amount of ti unit time to get it well-roasted, and want to get them before time ei(Just at exactly time ei is also OK). The roaster has a big grill which can hold an unlimited amount of kebabs (Unbelievable huh? Trust me, it’s real!). But he has so little charcoal that at most M kebabs can be roasted at the same time. He is skillful enough to take no time changing the kebabs being roasted. Can you help him determine if he can meet all the customers’ demand?
Oh, I forgot to say that the roaster needs not to roast a single kebab in a successive period of time. That means he can divide the whole ti unit time into k (1<=k<=ti) parts such that any two adjacent parts don’t have to be successive in time. He can also divide a single kebab into k (1<=k<=ti) parts and roast them simultaneously. The time needed to roast one part of the kebab well is linear to the amount of meat it contains. So if a kebab needs 10 unit time to roast well, he can divide it into 10 parts and roast them simultaneously just one unit time. Remember, however, a single unit time is indivisible and the kebab can only be divided into such parts that each needs an integral unit time to roast well.
Now N customers is coming. Customer i will arrive at time si (which means the roaster cannot serve customer i until time si). He/She will order ni kebabs, each one of which requires a total amount of ti unit time to get it well-roasted, and want to get them before time ei(Just at exactly time ei is also OK). The roaster has a big grill which can hold an unlimited amount of kebabs (Unbelievable huh? Trust me, it’s real!). But he has so little charcoal that at most M kebabs can be roasted at the same time. He is skillful enough to take no time changing the kebabs being roasted. Can you help him determine if he can meet all the customers’ demand?
Oh, I forgot to say that the roaster needs not to roast a single kebab in a successive period of time. That means he can divide the whole ti unit time into k (1<=k<=ti) parts such that any two adjacent parts don’t have to be successive in time. He can also divide a single kebab into k (1<=k<=ti) parts and roast them simultaneously. The time needed to roast one part of the kebab well is linear to the amount of meat it contains. So if a kebab needs 10 unit time to roast well, he can divide it into 10 parts and roast them simultaneously just one unit time. Remember, however, a single unit time is indivisible and the kebab can only be divided into such parts that each needs an integral unit time to roast well.
题意描述：有n个人来烤肉店吃烤肉，每个人在si 时刻来ei 时刻离开并且点了ni 份，每份烤肉要烤到ti 个单位时间才算烤熟，烤肉店里可以同时烤m份。问是否有一种计划使得n个人都可以拿到自己的ni 份。
算法分析：这道题本身不是很难，网络流的模型也很常见，但是这道题中（si，ei）的时间跨度很大（1<=si<=ei<=1000000），所以不能把时间区间直接拆分开建立模型，这样顶点个数太多，会超时。这里，介绍一下学到的新技巧，我们可以把时间区间压缩：time[]里保存全部的si 和 ei ，这样time[i]-time[i-1]就表示一段时间区间了，如果这段时间区间在[si，ei]中，那么就把第 i 个人这个顶点和第 i 个时间区间相连接，然后，from->第 i 个人，第 i 个时间区间->to。这样，每一条s-t路径就表示一种烤肉的时间计划了。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #include<cmath> 6 #include<algorithm> 7 #include<queue> 8 #include<vector> 9 #define inf 0x7fffffff 10 using namespace std; 11 const int maxn=600+10; 12 const int M = 999999; 13 14 int n,m,from,to; 15 int d[maxn]; 16 struct node 17 { 18 int v,flow; 19 int next; 20 }edge[M*2]; 21 int head[maxn],edgenum; 22 23 void add(int u,int v,int flow) 24 { 25 edge[edgenum].v=v ;edge[edgenum].flow=flow ; 26 edge[edgenum].next=head[u]; 27 head[u]=edgenum++; 28 29 edge[edgenum].v=u ;edge[edgenum].flow=0; 30 edge[edgenum].next=head[v]; 31 head[v]=edgenum++; 32 } 33 34 int bfs() 35 { 36 memset(d,0,sizeof(d)); 37 d[from]=1; 38 queue<int> Q; 39 Q.push(from); 40 while (!Q.empty()) 41 { 42 int u=Q.front() ;Q.pop() ; 43 for (int i=head[u] ;i!=-1 ;i=edge[i].next) 44 { 45 int v=edge[i].v; 46 if (!d[v] && edge[i].flow>0) 47 { 48 d[v]=d[u]+1; 49 Q.push(v); 50 if (v==to) return 1; 51 } 52 } 53 } 54 return 0; 55 } 56 57 int dfs(int u,int flow) 58 { 59 if (u==to || flow==0) return flow; 60 int cap=flow; 61 for (int i=head[u] ;i!=-1 ;i=edge[i].next) 62 { 63 int v=edge[i].v; 64 if (d[v]==d[u]+1 && edge[i].flow>0) 65 { 66 int x=dfs(v,min(cap,edge[i].flow)); 67 edge[i].flow -= x; 68 edge[i^1].flow += x; 69 cap -= x; 70 if (cap==0) return flow; 71 } 72 } 73 return flow-cap; 74 } 75 76 int dinic() 77 { 78 int sum=0; 79 while (bfs()) sum += dfs(from,inf); 80 return sum; 81 } 82 83 int main() 84 { 85 while (scanf("%d%d",&n,&m)!=EOF) 86 { 87 memset(head,-1,sizeof(head)); 88 edgenum=0; 89 int s[222],q[222],e[222],t[222]; 90 int time[maxn],cnt=1; 91 memset(time,0,sizeof(time)); 92 int sum=0; 93 for (int i=1 ;i<=n ;i++) 94 { 95 scanf("%d%d%d%d",&s[i],&q[i],&e[i],&t[i]); 96 sum += q[i]*t[i]; 97 time[cnt++]=s[i]; 98 time[cnt++]=e[i]; 99 } 100 sort(time+1,time+cnt); 101 int c=0; 102 for (int i=1 ;i<cnt ;i++) 103 { 104 if (time[c] != time[i]) 105 time[++c]=time[i]; 106 } 107 from=n+c+1; 108 to=from+1; 109 for (int i=1 ;i<=n ;i++) 110 add(from,i,q[i]*t[i]); 111 for (int i=1 ;i<=c ;i++) 112 { 113 add(n+i,to,m*(time[i]-time[i-1])); 114 for (int j=1 ;j<=n ;j++) 115 { 116 if (s[j]<=time[i-1] && time[i]<=e[j]) 117 add(j,n+i,inf); 118 } 119 } 120 if (sum==dinic()) printf("Yes "); 121 else printf("No "); 122 } 123 return 0; 124 }