POJ 3680 最小费用最大流

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
const int maxn = 500;
const int maxm = 1E5 + 10;
const int INF = 0x3f3f3f3f;
struct Edge
{
int to, next, cap, flow, cost;
};
Edge edge[maxm];
int head[maxn], tol, pre[maxn], dis[maxn], n, k, m, a[maxn], b[maxn], w[maxn], res[maxm], T, cnt;
bool vis[maxn];
void addedge(int u, int v, int cost, int cap)
{
edge[tol].to = v;
edge[tol].cap = cap;
edge[tol].cost = cost;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = 0;
edge[tol].cost = -cost;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}
bool spfa(int s, int t)
{
queue<int>q;
memset(dis, INF, sizeof(dis));
memset(vis, 0, sizeof(vis));
memset(pre, -1, sizeof(pre));
dis[s] = 0;
vis[s] = true;
q.push(s);
while (!q.empty())
{
int u = q.front(); q.pop();
vis[u] = false;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost)
{
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if (!vis[v])
{
vis[v] = true;
q.push(v);
}
}
}
}
return (pre[t] == -1 ? false : true);
}
int MinCostMaxFlow(int s, int t, int &cost)
{
int flow = 0; cost = 0;
while (spfa(s, t))
{
int Min = INF;
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
Min = min(Min, edge[i].cap - edge[i].flow);
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
{
edge[i].flow += Min;
edge[i ^ 1].flow -= Min;
cost += edge[i].cost * Min;
}
flow += Min;
}
return flow;
}
int main(int argc, char const *argv[])
{
scanf("%d", &T);
while (T--)
{
cnt = 0, tol = 0;
memset(head, -1, sizeof(head));
memset(res, -1, sizeof(res));
scanf("%d%d", &n, &k);
for (int i = 0; i < n; i++)
scanf("%d%d%d", &a[i], &b[i], &w[i]), res[a[i]] = 0, res[b[i]] = 0;
for (int i = 0; i < maxm; i++)
if (res[i] == 0) res[i] = ++cnt;
for (int i = 0; i < cnt; i++)
addedge(i + 1, i + 2, 0, INF);
for (int i = 0; i < n; i++)
addedge(res[a[i]], res[b[i]], -w[i], 1);
addedge(0, 1, 0, k);
addedge(cnt, cnt + 1, 0, k);
int ans = 0;
MinCostMaxFlow(0, cnt + 1, ans);
printf("%d\n", -ans);
}
return 0;
}

题意：给定n个带权开区间，选择其中一些使得权值最大并且区间重叠层数不超过k。

题解：最小费用流，区间有两百个，可以用左边的点发出一条到右边的点的边，容量为1，费用为负的权值。然后从左往右将依次将相邻的两个点都连起来，权值为0，容量为k，也就是说，如果选了这个区间，就会从费用为负数的边流过去，否则，就是从这个费用为0的边流过去。然后建立一个虚拟源点与最左边的点相连，权值为0，容量为k，这样就保证了重叠数之多为k，因为增广路上所经过的区间必定是不重合的，而流量只有k，所以满足题意。