POJ 3762 The Bonus Salary!（費用流）

題目鏈接：http://poj.org/problem?id=3762

 

題意：有N個任務，每個任務有開始時間，結束時間和價值，現在有M天，每天的同一個時刻只能執行一個任務且每個任務必須連續執行，問能獲得的最大價值

 

思路：先將每個任務的開始時間和結束時間先hash後離散，這樣就可使點集數目cnt在2 * N以內了，題目也就相當於給出了若干區間段，選出權值和儘量大的一些區間使得任意一個數最多被M個區間所覆蓋，也就是經典的區間K覆蓋問題了，建圖如下：

 

1.源點到1結點連邊容量爲K費用爲0

2.i 到 i + 1結點連邊容量爲K費用爲0

3.對於每個區間[L, R]，L到R結點連邊容量爲1費用爲-w

4.cnt向匯點連邊容量爲K費用爲0

 

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <climits>
#include <functional>
#include <deque>
#include <ctime>
#include <string>

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#pragma comment(linker, "/STACK:102400000,102400000")

using namespace std;

const int MAXN = 6000;
const int MAXM = 500000;
const int INF = 0x3f3f3f3f;

typedef long long ll;

struct Edge
{
	int to, next, cap, flow, cost;
} edge[MAXM];

int head[MAXN], tol;
int pre[MAXN], dis[MAXN];
bool vis[MAXN];
int N;//節點總個數，節點編號從0~N-1

void init(int n)
{
	N = n;
	tol = 0;
	memset(head, -1, sizeof(head));
}

void addedge(int u, int v, int cap, int cost)
{
	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;
	for (int i = 0; i < N; i++)
	{
		dis[i] = INF;
		vis[i] = false;
		pre[i] = -1;
	}
	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);
				}
			}
		}
	}
	if (pre[t] == -1) return false;
	else return true;
}

int res = 0;

//返回的是最大流，cost存的是最小費用
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])
		{
			if (Min > edge[i].cap - edge[i].flow)
				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;
			res = min(cost, res);
		}
		flow += Min;
	}
	return flow;
}

struct node
{
	int s, t, val;

	bool operator < (const node &rhs) const
	{
		if (s != rhs.s) return s < rhs.s;
		return t < rhs.t;
	}
} p[MAXN];

int hash(char *s)
{
	int s1 = (s[0] - '0') * 10 + (s[1] - '0');
	int s2 = (s[3] - '0') * 10 + (s[4] - '0');
	int s3 = (s[6] - '0') * 10 + (s[7] - '0');
	return s1 * 3600 + s2 * 60 + s3;
}

int ti[MAXN];

int main()
{
	int n, k;
	while (~scanf("%d%d", &n, &k))
	{
		int cnt = 0;
		for (int i = 0; i < n; i++)
		{
			char s[15];
			scanf("%s", s);
			p[i].s = hash(s);
			ti[cnt++] = p[i].s;

			scanf("%s", s);
			p[i].t = hash(s);
			ti[cnt++] = p[i].t;

			int w;
			scanf("%d", &w);
			p[i].val = w;
		}

		sort(ti, ti + cnt);
		cnt = unique(ti, ti + cnt) - ti;
		for (int i = 0; i < n; i++)
		{
			p[i].s = lower_bound(ti, ti + cnt, p[i].s) - ti + 1;
			p[i].t = lower_bound(ti, ti + cnt, p[i].t) - ti + 1;
		}

		// for (int i = 0; i < n; i++)
		// 	printf("%d %d %d\n", p[i].s, p[i].t, p[i].val);

		int s = 0, t = cnt + 1;
		init(t + 1);

		addedge(s, 1, k, 0);
		addedge(cnt, t, k, 0);
		for (int i = 0; i < n; i++)
			addedge(p[i].s, p[i].t, 1, -p[i].val);
		for (int i = 1; i < cnt; i++)
			addedge(i, i + 1, k, 0);

		res = 0;
		int ans;
		minCostMaxflow(s, t, ans);
		cout << -res << endl;
	}
	return 0;
}