UVaLive 2796 Concert Hall Scheduling (最小费用流)
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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; }
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