poj3680

题解:

相邻的建边

每一段建边

然后见一个原点,汇点

代码:

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int N=;
int b[N],T,c[N],fi[N],n,t,cas,m,x,mp[N],k,a[N];
int y,z,f[N],ne[N],num,zz[N],fl[N],gp[N],dist[N],pre[N],sl[N];
void jb(int x,int y,int z,int s)
{
ne[num]=fi[x];
fi[x]=num;
zz[num]=y;
sl[num]=z;
fl[num++]=s;
ne[num]=fi[y];
fi[y]=num;
zz[num]=x;
sl[num]=;
fl[num++]=-s;
}
int spfa()
{
memset(dist,0x3f,sizeof dist);
memset(pre,-,sizeof pre);
memset(gp,,sizeof gp);
memset(f,,sizeof f);
queue<int > Q;
Q.push();
dist[]=;
while (!Q.empty())
{
int now=Q.front();
Q.pop();
f[now]=;
for (int i=fi[now];i!=-;i=ne[i])
if (sl[i]>)
{
int t=zz[i];
if (dist[t]>dist[now]+fl[i])
{
dist[t]=dist[now]+fl[i];
pre[t]=now;
gp[t]=i;
if (!f[t])
{
f[t]=;
Q.push(t);
}
}
}
}
if (pre[n+]==-)return ;
return ;
}
void Max_flow()
{
int cost=,flow=;
while (!spfa())
{
int f=1e9;
for (int i=n+;i>;i=pre[i])
f=min(f,sl[gp[i]]);
cost+=f;
flow+=dist[n+]*f;
for (int i=n+;i>;i=pre[i])
{
sl[gp[i]]-=f;
sl[gp[i]^]+=f;
}
}
printf("%d\n",-flow);
}
void init()
{
// printf("Instance #%d: ",++cas);
num=;
memset(mp,,sizeof mp);
memset(fi,-,sizeof fi);
}
void doit()
{
scanf("%d%d",&n,&k);
for (int i=;i<=n;i++)
{
scanf("%d%d%d",&a[i],&b[i],&c[i]);
f[i*-]=a[i];f[i*]=b[i];
}
sort(f+,f+n+n+);
mp[f[]]=;
int tot=;
for (int i=;i<=*n;i++)
if (f[i]!=f[tot])mp[f[i]]=++tot,f[tot]=f[i];
for (int i=;i<=tot;i++)jb(i,i+,k,);
for (int i=;i<=n;i++)jb(mp[a[i]],mp[b[i]],,-c[i]);
jb(,,k,);jb(tot+,tot+,k,);
n=tot;
Max_flow();
}
int main()
{
scanf("%d",&T);
while (T--)
{
init();
doit();
}
}
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