UVaLive 2796 Concert Hall Scheduling (最小费用流)

题意：个著名的音乐厅因为财务状况恶化快要破产，你临危受命，试图通过管理的手段来拯救它，方法之一就是优化演出安排，既聪明的决定接受或拒绝哪些乐团的演出申请，使得音乐厅的收益最大化。该音乐厅有两个完全相同的房间，因此个乐团在申请演出的时候并不会指定房间，你只需要随便分配一个即可。每个演出都会持续若干天，每个房间每天只能举行一场演出。申请数目n为不超过100的正整数，每个申请用3个整数i,j,w来表示，表示从第i天到第j天，愿意支付w元。

析：把每一天都看成是一个结点，然后相邻两天加一个容量为2，费用为0的边，然后对于每个区间可以直接从左端点到右端点+1连一条容量为1，费用为-w的边，最后从最左边跑到最右边一次最小费用流量，取反即可。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-3;
const int maxn = 400 + 10;
const int maxm = 3e5 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

struct Edge{
  int from, to, cap, flow, cost;
};

struct MinCostMaxFlow{
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[maxn];
  int d[maxn];
  int p[maxn];
  bool inq[maxn];
  int a[maxn];

  void init(int n){
    this-> n = n;
    for(int i = 0; i < n; ++i)  G[i].cl;
    edges.cl;
  }

  void addEdge(int from, int to, int cap, int cost){
    edges.pb((Edge){from, to, cap, 0, cost});
    edges.pb((Edge){to, from, 0, 0, -cost});
    m = edges.sz;
    G[from].pb(m - 2);
    G[to].pb(m - 1);
  }

  bool bellman(int &flow, int &cost){
    ms(inq, 0);  ms(d, INF);  inq[s] = 1;
    d[s] = 0;  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].sz; ++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;
          a[e.to] = min(a[u], e.cap - e.flow);
          p[e.to] = G[u][i];
          if(!inq[e.to]){ inq[e.to] = 1;  q.push(e.to); }
        }
      }
    }
    if(d[t] == INF)  return false;
    flow += a[t];
    cost += d[t] * a[t];
    int u = t;
    while(u != s){
      edges[p[u]].flow += a[t];
      edges[p[u]^1].flow -= a[t];
      u = edges[p[u]].from;
    }
    return true;
  }

  int mincostmaxflow(int s, int t, int &flow){
    this-> s = s;
    this-> t = t;
    int cost = 0;
    while(bellman(flow, cost));
    return cost;
  }
};

MinCostMaxFlow mcmf;

int main(){
  while(scanf("%d", &n) == 1 && n){
    int s = 0, t = 366;
    mcmf.init(t + 5);
    for(int i = 0; i <= 365; ++i)  mcmf.addEdge(i, i+1, 2, 0);
    while(n--){
      int u, v, c;
      scanf("%d %d %d", &u, &v, &c);
      mcmf.addEdge(u, v+1, 1, -c);
    }
    int flow = 0;
    printf("%d\n", -mcmf.mincostmaxflow(s, t, flow));
  }
  return 0;
}