HDU

来源：互联网发布：手机共享网络给电脑编辑：程序博客网时间：2024/06/03 20:00
和上一道题差不多，都是时间的最大流。时间段的取值很大，所以加一个t数组离散化一下。
#include<bits/stdc++.h>using namespace std;const int N=1005;const int M=100010;const int INF=0x3f3f3f3f;struct node{    int u,v,cap,next;}eage[M];int head[N];int cur[N],dep[N],gap[N];int S[N];int source,sink;int top;int n,m;int ss[N],nn[N],ee[N],tt[N];int t[N*10];void Add(int u,int v,int w){    eage[top].u=u;    eage[top].v=v;    eage[top].cap=w;    eage[top].next=head[u];    head[u]=top++;    eage[top].u=v;    eage[top].v=u;    eage[top].cap=0;    eage[top].next=head[v];    head[v]=top++;}void BFS(){    queue<int>q;    memset(dep,-1,sizeof(dep));    memset(gap,0,sizeof(gap));    gap[0]=1;    dep[sink]=0;    q.push(sink);    while(!q.empty())    {        int u=q.front();        q.pop();        for(int i=head[u];i!=-1;i=eage[i].next)        {            int v=eage[i].v;            if(dep[v]==-1)            {                q.push(v);                dep[v]=dep[u]+1;                gap[dep[v]]++;            }        }    }}int Sap(){    BFS();    memcpy(cur,head,sizeof(head));    int tot=0;    int u=source;    int ans=0;    while(dep[source]<n)    {        if(u==sink)        {            int Min=INF;            int inser;            for(int i=0;i<=tot-1;i++)            {                if(Min>eage[S[i]].cap)                {                    Min=eage[S[i]].cap;                    inser=i;                }            }            for(int i=0;i<=tot-1;i++)            {                eage[S[i]].cap-=Min;                eage[S[i]^1].cap+=Min;            }            ans+=Min;            tot=inser;            u=eage[S[tot]].u;        }        if(u!=sink&&gap[dep[u]-1]==0)          break;        int v;        int i;        for(i=cur[u];i!=-1;i=eage[i].next)        {            v=eage[i].v;            if(eage[i].cap&&dep[v]+1==dep[u])            {                break;            }        }        if(i!=-1)        {            cur[u]=i;            S[tot++]=i;            u=v;            continue;        }        int Min=n;        for(int i=head[u];i!=-1;i=eage[i].next)        {            int v=eage[i].v;            if(eage[i].cap&&dep[v]<Min)            {                Min=dep[v];                cur[u]=i;            }        }        gap[dep[u]]--;        dep[u]=Min+1;        gap[dep[u]]++;        if(u!=source)u=eage[S[--tot]].u;    }    return ans;}int main(){    while(~scanf("%d%d",&n,&m))    {        memset(head,-1,sizeof(head));        top=0;        int sum=0;        source=0;        int o=0;        for(int i=1;i<=n;i++)        {            scanf("%d%d%d%d",&ss[i],&nn[i],&ee[i],&tt[i]);            sum+=nn[i]*tt[i];            Add(source,i,nn[i]*tt[i]);            t[o++]=ss[i];            t[o++]=ee[i];        }        sort(t,t+o);        int cnt=unique(t,t+o)-t;        sink=n+cnt+1;        for(int i=1;i<=cnt-1;i++)        {            Add(n+i,sink,m*(t[i]-t[i-1]));            for(int j=1;j<=n;j++)            {                if(ss[j]<=t[i-1]&&t[i]<=ee[j])                {                    Add(j,n+i,INF);                }            }        }        n=n+cnt+2;        if(sum==Sap())printf("Yes\n");        else printf("No\n");    }    return 0;}
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