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// Ladders and Snakes
// Solution by Jacob Plachta
#define DEBUG 0
#include <algorithm>
#include <functional>
#include <numeric>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <complex>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <sstream>
using namespace std;
#define LL long long
#define LD long double
#define PR pair<int,int>
#define Fox(i,n) for (i=0; i<n; i++)
#define Fox1(i,n) for (i=1; i<=n; i++)
#define FoxI(i,a,b) for (i=a; i<=b; i++)
#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
#define FoxR1(i,n) for (i=n; i>0; i--)
#define FoxRI(i,a,b) for (i=b; i>=a; i--)
#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
#define Min(a,b) a=min(a,b)
#define Max(a,b) a=max(a,b)
#define Sz(s) int((s).size())
#define All(s) (s).begin(),(s).end()
#define Fill(s,v) memset(s,v,sizeof(s))
#define pb push_back
#define mp make_pair
#define x first
#define y second
template<typename T> T Abs(T x) { return(x<0 ? -x : x); }
template<typename T> T Sqr(T x) { return(x*x); }
string plural(string s) { return(Sz(s) && s[Sz(s)-1]=='x' ? s+"en" : s+"s"); }
const int INF = (int)1e9;
const LD EPS = 1e-12;
const LD PI = acos(-1.0);
#if DEBUG
#define GETCHAR getchar
#else
#define GETCHAR getchar_unlocked
#endif
bool Read(int &x)
{
char c,r=0,n=0;
x=0;
for(;;)
{
c=GETCHAR();
if ((c<0) && (!r))
return(0);
if ((c=='-') && (!r))
n=1;
else
if ((c>='0') && (c<='9'))
x=x*10+c-'0',r=1;
else
if (r)
break;
}
if (n)
x=-x;
return(1);
}
#define LIM 50
struct Dinic {
struct Edge {
int to, rev, c, f;
Edge(int to, int rev, int c, int f) : to(to), rev(rev), c(c), f(f) {}
};
vector<int> lvl, ptr, q;
vector< vector<Edge> > adj;
Dinic(int n) : lvl(n), ptr(n), q(n), adj(n) {}
void addEdge(int a, int b, int c) {
adj[a].pb(Edge(b, Sz(adj[b]), c, 0));
adj[b].pb(Edge(a, Sz(adj[a]) - 1, 0, 0));
}
int dfs(int v, int t, int f) {
if (v == t || !f) return f;
for (int& i = ptr[v]; i < Sz(adj[v]); i++) {
Edge& e = adj[v][i];
if (lvl[e.to] == lvl[v] + 1) {
if (int p = dfs(e.to, t, min(f, e.c - e.f))) {
e.f += p, adj[e.to][e.rev].f -= p;
return p;
}
}
}
return 0;
}
int calc(int s, int t) {
int L, flow = 0; q[0] = s;
Fox(L,31) {
do {
lvl = ptr = vector<int>(Sz(q));
int qi = 0, qe = lvl[s] = 1;
while (qi < qe && !lvl[t]) {
int i, v = q[qi++];
Fox(i,Sz(adj[v])) {
Edge e=adj[v][i];
if (!lvl[e.to] && (e.c - e.f) >> (30 - L)) {
q[qe++] = e.to, lvl[e.to] = lvl[v] + 1;
}
}
}
while (int p = dfs(s, t, INF)) flow += p;
} while (lvl[t]);
}
return flow;
}
};
int main()
{
if (DEBUG)
freopen("in.txt","r",stdin);
// vars
int T,t;
int N,H;
int i,j,k;
int X[LIM],A[LIM],B[LIM];
// testcase loop
Read(T);
Fox1(t,T)
{
// input
Read(N),Read(H);
Fox(i,N)
Read(X[i]),Read(A[i]),Read(B[i]);
// construct flow graph
Dinic D(N+2);
Fox(i,N)
{
// connected to bottom?
if (A[i]==0)
D.addEdge(N,i,1e7);
// connected to top?
if (B[i]==H)
D.addEdge(i,N+1,1e7);
// consider connections to other ladders
Fox(j,N)
if (X[i]<X[j])
{
// compute combined length of connecting, unobstructed y-coordinate intervals
vector<pair<int,PR> > E;
Fox(k,N)
if ((X[i]<=X[k]) && (X[k]<=X[j]))
{
E.pb(mp(A[k],mp(1,k)));
E.pb(mp(B[k],mp(0,k)));
}
int s=0;
set<int> S;
sort(All(E));
Fox(k,Sz(E))
{
if (E[k].y.x)
S.insert(E[k].y.y);
else
S.erase(E[k].y.y);
if ((Sz(S)==2) && (S.count(i)) && (S.count(j)))
s+=E[k+1].x-E[k].x;
}
if (s)
{
D.addEdge(i,j,s);
D.addEdge(j,i,s);
}
}
}
// compute min cut
int ans=D.calc(N,N+1);
if (ans>=1e7)
ans=-1;
// output
printf("Case #%d: %d\n",t,ans);
}
return(0);
} |