kebab
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Description
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.
Input
There are multiple test cases. The first line of each case contains two positive integers N and M. N is the number of customers and M is the maximum kebabs the grill can roast at the same time. Then follow N lines each describing one customer, containing four integers: si (arrival time), ni (demand for kebabs), ei (deadline) and ti (time needed for roasting one kebab well).
There is a blank line after each input block.
Restriction:
1 <= N <= 200, 1 <= M <= 1,000
1 <= ni, ti <= 50
1 <= si < ei <= 1,000,000
There is a blank line after each input block.
Restriction:
1 <= N <= 200, 1 <= M <= 1,000
1 <= ni, ti <= 50
1 <= si < ei <= 1,000,000
Output
If the roaster can satisfy all the customers, output “Yes” (without quotes). Otherwise, output “No”.
Sample Input
2 10
1 10 6 3
2 10 4 2
2 10
1 10 5 3
2 10 4 2
Sample Output
Yes
No
题意：就是有一些人来买烤肉串，在si时候到，在ei时候走，要定ni个，每个肉串要花ti时间制作，问你能不能满足所有人的订单。
分析：题目最后一段就是说你可以将肉串拆成更小的，每个需要1个时间制作，这样就好做了。
首先源点和顾客相连，容量为ni*ti，将每个顾客来的时间段看做一个节点，有重复的，所以要去重，排序就行了。然后每个时间节点
和汇点相连，容量为时间段长度*M。之后考察每个时间点，如果有个顾客来的时间包含这个时间节点，就将这个顾客和这个时间节点
相连，容量设为INF就行了。
#include <iostream> #include <string.h> #include <algorithm> #include <stdio.h> using namespace std; const int MAXN=1010; const int MAXM=200010; const int INF=0x3f3f3f3f; struct Node { int to,next,cap; }edge[MAXM]; int tol; int head[MAXN]; int gap[MAXN],dis[MAXN],pre[MAXN],cur[MAXN]; void init() { tol=0; memset(head,-1,sizeof(head)); } void addedge(int u,int v,int w,int rw=0) { edge[tol].to=v;edge[tol].cap=w;edge[tol].next=head[u];head[u]=tol++; edge[tol].to=u;edge[tol].cap=rw;edge[tol].next=head[v];head[v]=tol++; } int sap(int start,int end,int nodenum) { memset(dis,0,sizeof(dis)); memset(gap,0,sizeof(gap)); memcpy(cur,head,sizeof(head)); int u=pre[start]=start,maxflow=0,aug=-1; gap[0]=nodenum; while(dis[start]<nodenum) { loop: for(int &i=cur[u];i!=-1;i=edge[i].next) { int v=edge[i].to; if(edge[i].cap&&dis[u]==dis[v]+1) { if(aug==-1||aug>edge[i].cap) aug=edge[i].cap; pre[v]=u; u=v; if(v==end) { maxflow+=aug; for(u=pre[u];v!=start;v=u,u=pre[u]) { edge[cur[u]].cap-=aug; edge[cur[u]^1].cap+=aug; } aug=-1; } goto loop; } } int mindis=nodenum; for(int i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].to; if(edge[i].cap&&mindis>dis[v]) { cur[u]=i; mindis=dis[v]; } } if((--gap[dis[u]])==0)break; gap[dis[u]=mindis+1]++; u=pre[u]; } return maxflow; } int n[210],s[210],e[210],t[210]; int a[210]; int main() { int N,M; while(scanf("%d%d",&N,&M)!=EOF){ init(); int sum=0; int num=0; for(int i=1;i<=N;i++){ scanf("%d%d%d%d",&s[i],&n[i],&e[i],&t[i]); sum+=n[i]*t[i]; addedge(0,i,n[i]*t[i]); a[num++]=s[i]; a[num++]=e[i]; } sort(a,a+num); num=unique(a,a+num)-a; int st=0,ed=num+N,nodenum=num+N+1; for(int i=0;i<num-1;i++){ addedge(N+1+i,ed,(a[i+1]-a[i])*M); for(int j=1;j<=N;j++){ if(s[j]<=a[i]&&e[j]>=a[i+1]) addedge(j,N+1+i,INF); } } if(sap(st,ed,nodenum)==sum) printf("Yes "); else printf("No "); } return 0; }