POJ 3469 —— 最大流

Dual Core CPU

Time Limit: 15000MS		Memory Limit: 131072K
Total Submissions: 17034		Accepted: 7350
Case Time Limit: 5000MS

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 A_i andB_i. 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, A_i and B_i.
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

Source

POJ Monthly--2007.11.25, Zhou Dong

題意是有兩個CPU，A和B，有N個任務要完成，n[i]在A上完成的花費是A[i]，在B上完成的花費是B[i]，其中有M個關聯事件，如果a任務和b任務在同一個CPU上完成則不要額外花費，否則額外花c元，求完成所有工作的最小花費。

思路是建一個超級源點S，一個超級匯點T，權值分別爲A[i]，B[i]，關聯事件任務的結點之間建一條邊，然後用Dinic求最大流（圖的規模比較大用Dinic比較好）。

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>
using namespace std;
///#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 20000 + 50;
const int MAXS = 200000 + 50;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
const int inf = 1 << 30;
#define eps 1e-10
const long long MOD = 1000000000 + 7;
const int mod = 10007;
typedef long long LL;
const double PI = acos(-1.0);
//typedef double D;
typedef pair<int , int> pii;
typedef vector<int> vec;
typedef vector<vec> mat;

#define Bug(s) cout << "s = " << s << endl;
///#pragma comment(linker, "/STACK:102400000,102400000";
int N , M;
int A[MAXN] , B[MAXN];
int a[MAXS] , b[MAXS] , c[MAXS];
struct edge{int to , cap , rev;};
vector<edge>G[MAXN];
int level[MAXN];
int iter[MAXN];
void add_edge(int from , int to , int cap)
{
    G[from].push_back((edge){to , cap , G[to].size()});
    G[to].push_back((edge){from , 0 , G[from].size() - 1});
}
void bfs(int s)
{
    memset(level , -1 , sizeof(level));
    queue<int>que;
    level[s] = 0;
    que.push(s);
    while(!que.empty())
    {
        int v = que.front();que.pop();
        for(int i = 0 ; i < G[v].size() ; i++)
        {
            edge &e = G[v][i];
            if(e.cap > 0 && level[e.to] < 0)
            {
                level[e.to] = level[v] + 1;
                que.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++)
    {
        edge &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;
    for(;;)
    {
        bfs(s);
        if(level[t] < 0)return flow;
        memset(iter , 0 , sizeof(iter));
        int f;
        while((f = dfs(s , t , INF)) > 0)flow += f;
    }
}
void solve()
{
    int s = N , t = s + 1;
    for(int  i = 0 ; i < N ; i++)
    {
        add_edge(i , t , A[i]);
        add_edge(s , i , B[i]);
    }
    for(int i = 0 ; i < M ; i++)
    {
        add_edge(a[i] - 1 , b[i] - 1 , c[i]);
        add_edge(b[i] - 1 , a[i] - 1 , c[i]);
    }
    printf("%d\n" , max_flow(s , t));
}
int main()
{
    while(~scanf("%d%d" , &N , &M))
    {
        for(int i = 0 ; i <N ; i++)scanf("%d%d" , &A[i] , &B[i]);
        for(int i = 0 ; i < M;  i++)scanf("%d%d%d" , &a[i] , &b[i] , &c[i]);
        solve();
    }
    return 0;
}