POJ 3680 Intervals(经典费用流）

解题思路：

区间K覆盖问题：数轴上有一些带权值的区间，选出权和尽量大的一些区间，使得任意一个点最多被K个区间覆盖。

构图方法为：把每一个数作为一个节点，然后对于权值为W的区间[ u, v ]连一条边，容量为1，费用为-w，再对所有相邻

的点连边i -> i + 1，容量为K，费用为0；最后求最左端到最右端的最小费用最大流即可。如果数值范围太大，需要先进行离散化。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <cmath>
#define LL long long
using namespace std;
const int maxn = 500 + 10;
const int INF = 0x3f3f3f3f;
struct Edge
{
int from, to, cap, flow, cost;
Edge(int u, int v, int c, int f, int w) : from(u), to(v), cap(c), flow(f), cost(w) { }
};
int n;
vector<Edge>edges;
vector<int>G[maxn];
int inq[maxn];
int d[maxn];
int p[maxn];
int a[maxn];
void init()
{
for(int i=0;i<n;i++)
G[i].clear();
edges.clear();
}
void AddEdge(int from, int to, int cap, int cost)
{
edges.push_back(Edge(from,to,cap,0,cost));
edges.push_back(Edge(to,from,0,0,-cost));
int M = edges.size();
G[from].push_back(M-2);
G[to].push_back(M-1);
}
bool SPFA(int s, int t, int& flow, LL& cost)
{
for(int i=0;i<=n+1;i++)
d[i] = INF;
memset(inq, 0, sizeof(inq));
d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
queue<int>Q;
Q.push(s);
while(!Q.empty())
{
int u = Q.front();Q.pop();
inq[u] = 0;
for(int i=0;i<G[u].size();i++)
{
Edge& e = edges[G[u][i]];
if(e.cap > e.flow && d[e.to] > d[u] + e.cost)
{
d[e.to] = d[u] + e.cost;
p[e.to] = G[u][i];
a[e.to] = min(a[u], e.cap - e.flow);
if(!inq[e.to]){Q.push(e.to);inq[e.to] = 1;}
}
}
}
if(d[t] == INF) return false;
flow += a[t];
cost += (LL) d[t] * (LL) a[t];
for(int u=t;u!=s;u=edges[p[u]].from)
{
edges[p[u]].flow += a[t];
edges[p[u]^1].flow -= a[t];
}
return true;
}
int MincostMaxflow(int s, int t, LL& cost)
{
int flow = 0; cost = 0;
while(SPFA(s,t,flow,cost));
return flow;
}
struct Node
{
int a;
int p;
bool operator < (const Node &rhs)const
{
return a < rhs.a;
}
}li[maxn];
struct edge
{
int u, v,w;
}ee[maxn];
int main()
{
int T, N, K;
scanf("%d", &T);
while(T--)
{
scanf("%d%d", &N, &K);
int u, v, w;
for(int i=1;i<=N;i++)
{
scanf("%d%d%d", &u, &v, &w);
li[2*i-1].a = u;
li[2*i-1].p = -i;
li[2*i].a = v;
li[2*i].p= i;
ee[i].u=u;
ee[i].v=v;
ee[i].w=w;
}
sort(li+1, li + 1 +2* N);
int tmp = li[1].a,c = 1;
for(int i=1;i<=2*N;i++)
{
if(li[i].a != tmp)
{
c++;
tmp = li[i].a;
}
if(li[i].p > 0)
{
ee[li[i].p].v = c;
}
else
ee[-li[i].p].u = c;
}
n = c + 1;
init();
for(int i=0;i<c;i++)
AddEdge(i,i+1,K,0);
for(int i=1;i<=N;i++)
AddEdge(ee[i].u,ee[i].v,1,-ee[i].w);
LL cost = 0;
int ans = MincostMaxflow(0,c,cost);
printf("%lld\n",-cost);
}
return 0;
}