uva 1317 ,poj 2047 Concert Hall Scheduling
Description
You are appointed director of a famous concert hall, to save it from bankruptcy. The hall is very popular, and receives many requests to use its two fine rooms, but unfortunately the previous director was not very efficient, and it has been losing money for many years. The two rooms are of the same size and arrangement. Therefore, each applicant wishing to hold a concert asks for a room without specifying which. Each room can be used for only one concert per day.
In order to make more money, you have decided to abandon the previous fixed price policy, and rather let applicants specify the price they are ready to pay. Each application shall specify a period [i, j] and an asking price w, where i and j are respectively the first and last days of the period (1 <= i <= j <= 365), and w is a positive integer in yen, indicating the amount the applicant is willing to pay to use a room for the whole period.
You have received applications for the next year, and you should now choose the applications you will accept. Each application should be either accepted for its whole period or completely rejected. Each concert should use the same room during the whole applied period.
Considering the dire economic situation of the concert hall, artistic quality is to be ignored, and you should just try to maximize the total income for the whole year by accepting the most profitable applications.
Input
The input has multiple data sets, each starting with a line consisting of a single integer n, the number of applications in the data set. Then, it is followed by n lines, each of which represents one application with a period [i, j] and an asking price w yen in the following format.
i j w
A line containing a single zero indicates the end of the input.
The maximum number of applications in a data set is one thousand, and the maximum asking price is one million yen.
Output
For each data set, print a single line containing an integer, the maximum total income in yen for the data set.
Sample Input
4
1 2 10
2 3 10
3 3 10
1 3 10
6
1 20 1000
3 25 10000
5 15 5000
22 300 5500
10 295 9000
7 7 6000
8
32 251 2261
123 281 1339
211 235 5641
162 217 7273
22 139 7851
194 198 9190
119 274 878
122 173 8640
0
Sample Output
30
25500
38595
题目大意：一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。
解题思路：设置366个结点，表示365天（结点1到结点2表示第一天）。设置超级源点，连向第一个结点，容量为2（两个音乐厅），费用为0。设置超级汇点，使结点366连向它，容量为2，费用为0。然后，前365个结点，每个节点连向他的下一个节点，容量为2，费用为0，表示时间的流逝。最后对于每个申请，i到j + 1连一条边，容量为1，费用为负的w。跑费用流。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <queue>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 405;
const int M = 700;
int n, s, t;
int pre[N], inq[N];
ll a[N], d[N];
struct Edge{
int from, to;
ll cap, flow;
ll cos;
};
vector<Edge> edges;
vector<int> G[M];
void init() {
for (int i = 0; i < M; i++) G[i].clear();
edges.clear();
}
void addEdge(int from, int to, ll cap, ll cos) {
edges.push_back((Edge){from, to, cap, 0, cos});
edges.push_back((Edge){to, from, 0, 0, -cos});
int m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
int BF(int s, int t, ll& flow, ll& cost) {
queue<int> Q;
memset(inq, 0, sizeof(inq));
memset(a, 0, sizeof(a));
memset(pre, 0, sizeof(pre));
for (int i = 0; i < N; i++) d[i] = INF;
d[s] = 0;
a[s] = INF;
inq[s] = 1;
pre[s] = 0;
Q.push(s);
while (!Q.empty()) {
int u = Q.front(); Q.pop();
inq[u] = 0;
for (int i = 0; i < G[u].size(); i++) {
Edge &e = edges[G[u][i]];
if (e.cap > e.flow && d[e.to] > d[u] + e.cos) {
d[e.to] = d[u] + e.cos;
a[e.to] = min(a[u], e.cap - e.flow);
pre[e.to] = G[u][i];
if (!inq[e.to]) {
inq[e.to] = 1;
Q.push(e.to);
}
}
}
}
if (d[t] == INF) return 0;
flow += a[t];
cost += (ll)d[t] * (ll)a[t];
for (int u = t; u != s; u = edges[pre[u]].from) {
edges[pre[u]].flow += a[t];
edges[pre[u]^1].flow -= a[t];
}
return 1;
}
int MCMF(int s, int t, ll& cost) {
ll flow = 0;
cost = 0;
while (BF(s, t, flow, cost));
return flow;
}
void input() {
s = 0, t = 400;
int a, b, c;
for (int i = 1; i <= n; i++) {
scanf("%d %d %d", &a, &b, &c);
addEdge(a, b + 1, 1, -c);
}
addEdge(s, 1, 2, 0);
for (int i = 1; i <= 366; i++) {
addEdge(i, i + 1, 2, 0);
}
addEdge(366, t, 2, 0);
}
int main() {
while (scanf("%d", &n) != EOF) {
if (n == 0) break;
init();
input();
ll cost;
MCMF(s, t, cost);
printf("%lld\n", -cost);
}
return 0;
}
