POJ: Dual Core CPU

Description

As more and more computers are equipped with dual core CPU, SetagLilb, the Chief Technology Officer of TinySoft Corporation, decided to update their famous product — SWODNIW.

The routine consists of N modules, and each of them should run in a certain core. The costs for all the routines to execute on two cores has been estimated. Let’s define them as Ai and Bi. Meanwhile, M pairs of modules need to do some data-exchange. If they are running on the same core, then the cost of this action can be ignored. Otherwise, some extra cost are needed. You should arrange wisely to minimize the total cost.

Input

There are two integers in the first line of input data, N and M (1 ≤ N ≤ 20000, 1 ≤ M ≤ 200000) .
The next N lines, each contains two integer, Ai and Bi.
In the following M lines, each contains three integers: a, b, w. The meaning is that if module a and module b don’t execute on the same core, you should pay extra w dollars for the data-exchange between them.

Output

Output only one integer, the minimum total cost.

Sample Input

3 1
1 10
2 10
10 3
2 3 1000

Sample Output

13

Problem Analysis:

We can convert the minimum costs of 2 sets problem into a max flow min cut problem.

First, we need to construct a cost graph. we create 2 virtual vertices s and t. Connect all routines to s and t, add edges for extra cost between routines.

Find the min cut of the graph, we can find it is the minimize total cost

Solution

#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
struct SEdge {
  int to;
  int cap;
  int rev;
  SEdge(int t, int c, int r): to(t), cap(c), rev(r) {}
};
#define MAX_N 20000
#define MAX_M 200000
#define INF 0x3f3f3f3f
vector<SEdge> G[MAX_N + 2];
int level[MAX_N + 2];
int iter[MAX_N + 2];
void add_edge(int from, int to, int cap) {
  SEdge fromEdge(to, cap, (int)G[to].size());
  G[from].push_back(fromEdge);
  SEdge toEdge(from, 0, (int)G[from].size() - 1);
  G[to].push_back(toEdge);
}
void bfs(int s) {
  memset(level, -1, sizeof(level));
  queue<int> q;
  level[s] = 0;
  q.push(s);
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < G[v].size(); ++i) {
      SEdge &e = G[v][i];
      if (e.cap > 0 && level[e.to] < 0) {
        level[e.to] = level[v] + 1;
        q.push(e.to);
      }
    }
  }
}
int dfs(int v, int t, int f) {
  if (v == t) return f;
  for (int &i = iter[v]; i < G[v].size(); ++i) {
    SEdge &e = G[v][i];
    if (e.cap > 0 && level[v] < level[e.to]) {
      int d = dfs(e.to, t, min(f, e.cap));
      if (d > 0) {
        e.cap -= d;
        G[e.to][e.rev].cap += d;
        return d;
      }
    }
  }
  return 0;
}
int max_flow(int s, int t) {
  int flow = 0;
  while (true) {
    bfs(s);
    if (level[t] < 0) return flow;
    memset(iter, 0, sizeof(iter));
    int f;
    while ((f = dfs(s, t, INF)) > 0) flow += f;
  }
}
int main(int argc, char *argv[]) {
  int n, m;
  scanf("%d%d", &n, &m);
  int s = n;
  int t = s + 1;
  for (int i = 0; i < n; ++i) {
    getchar();
    int A, B;
    scanf("%d%d", &A, &B);
    add_edge(i, t, A);
    add_edge(s, i, B);
  }
  for (int i = 0; i < m; ++i) {
    getchar();
    int a, b, w;
    scanf("%d%d%d", &a, &b, &w);
    add_edge(a - 1, b - 1, w);
    add_edge(b - 1, a - 1, w);
  }
  printf("%d\n", max_flow(s, t));
  return 0;
}

time complexity is ford-fulkerson algorithm O(EV²)