Dual Core CPU

描述

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.

输入

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.

输出

Output only one integer, the minimum total cost.

样例输入

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3 1
1 10
2 10
10 3
2 3 1000

样例输出

1
13

代码

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#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f;

struct edge {
    int dest;
    int capacity;
} edge_set[100000000];
vector<int> source[200000];
int n, m;
int level[200000];
int edge_count = 0;

void addedge(int u, int v, int capacity)
{
    edge_set[edge_count].dest = v;
    edge_set[edge_count].capacity = capacity;
    source[u].push_back(edge_count++);
    edge_set[edge_count].dest = u;
    edge_set[edge_count].capacity = 0;
    source[v].push_back(edge_count++);
}

bool bfs()
{
    memset(level, 0, sizeof(level));
    queue<int> bfs_queue;
    level[0] = 1;
    bfs_queue.push(0);
    while (!bfs_queue.empty()) {
        int cur = bfs_queue.front();
        bfs_queue.pop();
        if (cur == n + 1) {
            return true;
        }
        for (int i = 0; i < source[cur].size(); i++) {
            int dest = edge_set[source[cur][i]].dest;
            if (level[dest] == 0 && edge_set[source[cur][i]].capacity > 0) {
                level[dest] = level[cur] + 1;
                bfs_queue.push(dest);
            }
        }
    }
    return false;
}

int dfs(int s, int t, int max_f)
{
    if (s == t)
        return max_f;
    int ret = 0;
    for (int i = 0; i < source[s].size(); i++) {
        int src = source[s][i];
        int v = edge_set[src].dest, f = edge_set[src].capacity;
        if (level[s] + 1 == level[v] && f > 0) {
            int min_n = min(max_f - ret, f);
            f = dfs(v, t, min_n);
            if (src % 2 == 0) {
                edge_set[src].capacity -= f;
                edge_set[src + 1].capacity += f;
            }
            else {
                edge_set[src].capacity -= f;
                edge_set[src - 1].capacity += f;
            }
            ret += f;
            if (ret == max_f)
                return ret;
        }
    }
    return ret;
}

int main(void)
{
    scanf("%d%d", &n, &m);
    int a, b, c;
    for (int i = 1; i <= n; i++) {
        scanf("%d%d", &a, &b);
        addedge(0, i, a);
        addedge(i, n + 1, b);
    }
    for (int i = 0; i < m; i++) {
        scanf("%d%d%d", &a, &b, &c);
        addedge(a, b, c);
        addedge(b, a, c);
    }
    int max_flow = 0;
    // for(int i = 1; i <= n + 2; i ++){
    //     cout << "Source: " << i << endl;
    //     for(vector<int>::iterator it = source[i].begin(); it != source[i].end(); it++){
    //         cout << edge_set[*it].dest << '\t' << edge_set[*it].capacity << endl;
    //     }
    // }
    while (bfs()) {
        max_flow += dfs(0, n + 1, INF);
    }
    printf("%d\n", max_flow);
    return 0;
}

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