ZOJ3762 The Bonus Salary!(最小费用最大流)

题意：给你N个的任务一定要在每天的[Li,Ri]时段完成，然后你只有K天的时间，每个任务有个val，然后求K天里能够获得的最大bonus。

思路：拿到手第一直觉是最小费用最大流，然后不会建图，就跑去想dp去了。好吧最小费用最大流的做法是这样的。首先题目给的是时间，所以变成整数的区间再离散化是一个标准姿势啦。然后对于相邻的时间点 1，2，3，4顺次建立这样的边 1->2->3->4...，边的容量为inf,费用为0.然后对于源点s，我们向时间点1连一条容量为k，费用为0的边，目的是限流。最后由最后一个时间点向汇点t连一条容量为inf,费用为0的边。然后对于每个事件，li,ri,vi,由时间点li->ri连容量为1费用为-vi的边。跑一遍最大流就可以了，最后的答案取个相反数（因为这里其实是最大费用最大流）。考虑一下正确性，首先一定是可以满流的，然后对于每条流出的流量，它只能往右走不能往左走，保证了区间不相交，所以最后跑出来的就是答案。

#pragma warning(disable:4996)
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<cmath>
#include<queue>
#define ll long long
#define maxn 10000
#define maxe 30000
#define inf 0x3f3f3f3f
using namespace std;

struct Edge{
	int u, v, nxt, cap, cost;
}edge[maxe];
int head[maxn];

int n, m;
int k;
int siz;

struct MinCostMaxFlow
{
	queue<int> que;
	int add; // edges number
	int vn; // total vertex number
	int cost[maxn], in[maxn], pre[maxn];
	bool vis[maxn];
	void init(){
		add = 0; vn = siz + 10; memset(head, -1, sizeof(head));
		while (!que.empty()) que.pop();
	}
	void insert(int u, int v, int w, int c){
		edge[add].u = u; edge[add].v = v; edge[add].cap = w; edge[add].cost = c;
		edge[add].nxt = head[u]; head[u] = add++;
		edge[add].u = v; edge[add].v = u; edge[add].cap = 0; edge[add].cost = -c;
		edge[add].nxt = head[v]; head[v] = add++;
	}

	bool spfa(int s, int e){
		memset(cost, 0x3f3f3f3f, sizeof(cost));
		memset(in, 0, sizeof(in));
		memset(vis, 0, sizeof(vis));
		cost[s] = 0; pre[s] = -1;
		que.push(s); vis[s] = true; in[s]++;
		while (!que.empty()){
			int u = que.front(); que.pop();
			vis[u] = false;
			for (int i = head[u]; i != -1; i = edge[i].nxt){
				int v = edge[i].v;
				if (edge[i].cap > 0 && cost[v] > cost[u] + edge[i].cost){
					cost[v] = cost[u] + edge[i].cost; pre[v] = i;
					if (!vis[v]){
						que.push(v); vis[v] = true; in[v]++;
						if (in[v] > vn) return false;
					}
				}
			}
		}
		if (cost[e] < inf) return true;
		else return false;
	}
	int mincostmaxflow(int s, int e){
		int mincost = 0, maxflow = 0;
		while (spfa(s, e)){
			int flow = inf;
			for (int i = pre[e]; i != -1; i = pre[edge[i].u]){
				flow = min(flow, edge[i].cap);
			}
			maxflow += flow;
			for (int i = pre[e]; i != -1; i = pre[edge[i].u]){
				edge[i].cap -= flow;
				edge[i ^ 1].cap += flow;
			}
			mincost += cost[e] * flow;
		}
		return mincost;
	}
}net;

struct Seg
{
	int l, r;
	int val;
}seg[maxn];

int dis[maxn]; int top;

int main()
{
	while (cin >> n >> k){
		int hi, mi, si;
		top = 0;
		for (int i = 0; i < n; i++){
			scanf("%d:%d:%d", &hi, &mi, &si);
			seg[i].l = hi * 60*60 + mi * 60 + si;
			dis[top++] = seg[i].l;
			scanf("%d:%d:%d", &hi, &mi, &si);
			seg[i].r = hi * 60*60 + mi * 60 + si;
			dis[top++] = seg[i].r;
			scanf("%d", &seg[i].val);
		}
		sort(dis, dis + top);
		siz = unique(dis, dis + top) - dis;
		for (int i = 0; i < n; i++){
			seg[i].l = lower_bound(dis, dis + siz, seg[i].l) - dis + 1;
			seg[i].r = lower_bound(dis, dis + siz, seg[i].r) - dis + 1;
		}
		net.init();
		int s = siz + 1, t = s + 1;
		for (int i = 1; i <= siz - 1; i++) net.insert(i, i + 1, inf, 0);
		net.insert(s, 1, k, 0);
		net.insert(siz, t, inf, 0);
		for (int i = 0; i < n; i++){
			net.insert(seg[i].l, seg[i].r, 1, -seg[i].val);
		}
		cout << -net.mincostmaxflow(s, t) << endl;
	}
	return 0;
}