bzoj3572又TM是网络流

= =我承认我写网络流写疯了

= =我承认前面几篇博文都是扯淡,我写的是垃圾dinic(根本不叫dinic)

= =我承认这道题我调了半天

= =我承认我这道题一开始是T的,后来换上真正的dinic才过

= =我承认我还没理解dinic

= =我承认我一直到处改最后才发现输入顺序错了

= =我承认我已经疯了

= =我承认我写我承认上瘾了


回正题:这道题建图比较基础(但终于不是板题了),对于每个任务和每一天都建一个点(感觉把任务拆成若干点还能写二分图匹配),然后限制流量,然后判断是否满了

 1 #include <cstdio>
 2 #define INF 2147483647
 3 int n,m,N,need,p,q,o,zl,h,t,T,tt,ans,fir[1005],nex[600000],to[600000],flo[600000],d[1005],l[1005];
 4 inline int min(int p,int q){    return(p<q)?p:q;}
 5 inline void add(int p,int q,int o){    to[++N]=q;flo[N]=o;nex[N]=fir[p];fir[p]=N;
 6                                     to[++N]=p;flo[N]=0;nex[N]=fir[q];fir[q]=N;}
 7 int dfs(int now,int flow,int sum)
 8 {
 9     if(now==n+501) return flow;
10     for(int i=fir[now];i && (flow>sum);i=nex[i])
11         if(d[to[i]]==d[now]+1 && flo[i])
12             zl=dfs(to[i],min(flow,flo[i]),0),sum+=zl,flo[i]-=zl,flo[i^1]+=zl;
13     if(sum==0) d[now]=0;
14     return sum;    
15 }
16 bool bfs()
17 {
18     for(int i=1;i<=n+501;i++) d[i]=0;
19     for(h=1,t=1,l[1]=0,d[0]=1;h<=t;h++)
20         for(int i=fir[l[h]];i;i=nex[i])
21             if(!d[to[i]] && flo[i])
22                 l[++t]=to[i],d[l[t]]=d[l[h]]+1;
23     return d[n+501];
24 }
25 int main()
26 {
27     for(scanf("%d",&T),tt=1;tt<=T;tt++)
28     {
29         scanf("%d%d",&n,&m);need=0;N=1;
30         for(int i=0;i<=n+501;i++) fir[i]=0;
31         for(int i=1;i<=n;i++)
32         {
33             scanf("%d%d%d",&o,&p,&q),add(0,i,o),need+=o;
34             for(int j=p;j<=q;j++) add(i,j+n,1);
35         }
36         for(int i=1;i<=500;i++)
37             add(i+n,n+501,m);
38         for(ans=0;bfs();ans+=dfs(0,INF,0));
39         printf((ans==need)?"Case %d: Yes

":"Case %d: No

",tt);  
40     }
41     return 0;
42 }
原文地址:https://www.cnblogs.com/wanglichao/p/5818035.html