hdu2883

再来一题最大流。

这里的建模，与汇点连接的是时间区间。可以理解成在一段时间里总共可以制作多少个kebab。如何这段时间满足客人的要求，则可以将该区间与客人相连，弧的流量设为能够制作的上界，或者直接设为INF就行。其余的建模步骤和常规任务分配最大流建模方式类似。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#define INF 0x7fffffff
#define maxn 25000
using namespace std;
int n,m;
struct Edge{
int from,to,cap,flow;
};
class Dinic{
public:
int s,t,c;
vector<Edge>edges;
vector<int>G[maxn];
bool vis[maxn];
int dist[maxn];
int cur[maxn];
public:
void AddEdge(int from,int to,int cap){
edges.push_back((Edge){from,to,cap,0});
edges.push_back((Edge){to,from,0,0});
c=edges.size();
G[from].push_back(c-2);
G[to].push_back(c-1);
}
bool BFS(){
queue<int>Q;
memset(vis,0,sizeof(vis));
Q.push(s);
dist[s]=0;
vis[s]=1;
while(!Q.empty()){
int x=Q.front();Q.pop();
for(int i=0;i<G[x].size();i++){
Edge& e=edges[G[x][i]];
if(!vis[e.to]&&e.cap>e.flow){
vis[e.to]=1;
dist[e.to]=dist[x]+1;
Q.push(e.to);
}
}
}
return vis[t];
}
int DFS(int x,int a){
if(x==t||a==0)return a;
int flow=0,f;
for(int& i=cur[x];i<G[x].size();i++){
Edge& e=edges[G[x][i]];
if(dist[x]+1==dist[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0){
e.flow+=f;
edges[G[x][i]^1].flow-=f;
flow+=f;
a-=f;
if(a==0)break;
}
}
return flow;
}
int Maxflow(int s,int t){
this->s=s;this->t=t;
int flow=0;
while(BFS()){
memset(cur,0,sizeof(cur));
flow+=DFS(s,INF);
flow+=DFS(s,INF);
}
return flow;
}
void init(){
edges.clear();
for(int i=0;i<maxn;i++){
G[i].clear();
dist[i]=0;
}
}
}Do;
struct Time{
int s,e;
};
Time *timei;
int *timej;
int Judge(int i,int j){
if(timei[i].s>timej[j])
return 1;//不满足上界
if(timei[i].e<timej[j+1])
return 2;//不满足下界
return 3;//满足
}

int main(){
while(scanf("%d%d",&n,&m)!=EOF){
Do.init();
timei=new Time[n+1];
timej=new int[2*n+1];int pj=1,Sum=0;
for(int i=1;i<=n;i++){
int si,ni,ei,ti;
scanf("%d%d%d%d",&si,&ni,&ei,&ti);
Do.AddEdge(0,i,ni*ti);
Sum+=(ni*ti);
timei[i]=(Time){si,ei};
timej[pj++]=si;
timej[pj++]=ei;
}
sort(timej+1,timej+1+2*n);
for(int i=1;i<=n;i++){
for(int j=1;j<=n*2-1;j++){
if(Judge(i,j)==2)break;
if(Judge(i,j)==3)
Do.AddEdge(i,n+j,INF);
}
}
for(int j=1;j<=2*n-1;j++){
Do.AddEdge(n+j,3*n,(timej[j+1]-timej[j])*m);
}
if(Do.Maxflow(0,3*n)==Sum)printf("Yes\n");
else printf("No\n");
delete timei;
delete timej;
}
return 0;
}