POJ 2135 —— 最小费用流

Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9786   Accepted: 3603 Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <= 10000) paths that connect the fields in various ways. Each path connects two different fields and has a nonzero length smaller than 35,000.  To show off his farm in the best way, he walks a tour that starts at his house, potentially travels through some fields, and ends at the barn. Later, he returns (potentially through some fields) back to his house again.  He wants his tour to be as short as possible, however he doesn't want to walk on any given path more than once. Calculate the shortest tour possible. FJ is sure that some tour exists for any given farm. Input * Line 1: Two space-separated integers: N and M.  * Lines 2..M+1: Three space-separated integers that define a path: The starting field, the end field, and the path's length.  Output A single line containing the length of the shortest tour.  Sample Input 4 5 1 2 1 2 3 1 3 4 1 1 3 2 2 4 2 Sample Output 6 Source USACO 2003 February Green

题意是给你n块地，m条路，让你输出从1点出发经过所有点，回到1点的最短距离，保证每条道路只通过一次。

思路：我们可以把问题转化为从1到n，没有公共边的两条路径，相当于求流量为2的最小费用流。

#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 = 10000 + 50;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
const int MOD = (int)1e9 + 7;
typedef long long LL;
const double PI = acos(-1.0);

#define Bug(s) cout << "s = " << s << endl;
///#pragma comment(linker, "/STACK:102400000,102400000")

int n , m;
int a[MAXN] , b[MAXN] , c[MAXN];
typedef pair<int , int> pi;
///表示边的结构体（终点，容量，费用，反向边）
struct edge{int to , cap, cost , rev;};
int V;///顶点数
vector<edge> G[MAXN];///图的邻接表表示
int h[MAXN];///顶点的势
int dist[MAXN];///最短距离
int prevv[MAXN] , preve[MAXN];///最短路中的前驱节点和对应的边
///向图中增加一条从from到to容量为cap费用为cost的边
void add_edge(int from , int to , int cap , int cost)
{
    G[from].push_back((edge){to , cap ,cost , G[to].size()});
    G[to].push_back((edge){from , 0 , -cost , G[from].size() - 1});
}
///求解从s到t流量为f的最小费用流
///如果没有流量为f的流，则返回-1
int min_cost_flow(int s , int t , int f)
{
    int res = 0;
    fill(h , h + V ,0);///初始化h
    ///使用Dijkstra算法更新h
    while(f > 0)
    {
        priority_queue<pi , vector<pi> , greater<pi> > que;
        fill(dist , dist  + V , INF);
        dist[s] = 0;
        que.push(pi(0 , s));
        while(!que.empty())
        {
            pi p = que.top() ; que.pop();
            int v = p.second;
            if(dist[v] < p.first)continue;
            for(int i = 0 ; i < G[v].size() ; i++)
            {
                edge &e = G[v][i];
                if(e.cap > 0 && dist[e.to] > dist[v] + e.cost + h[v] - h[e.to])
                {
                    dist[e.to] = dist[v] + e.cost + h[v] - h[e.to];
                    prevv[e.to] = v;
                    preve[e.to] = i;
                    que.push(pi(dist[e.to] , e.to));
                }
            }
        }
        ///不能再增广
        if(dist[t] == INF)return -1;
        for(int v = 0 ; v < V; v++)h[v] += dist[v];
        ///沿s到t的最短路尽量增广
        int d = f;
        for(int v = t ; v != s ; v = prevv[v])
        {
            d = min(d , G[prevv[v]][preve[v]].cap);
        }
        f -= d;
        res += d * h[t];
        for(int v = t ; v != s ; v = prevv[v])
        {
            edge & e = G[prevv[v]][preve[v]];
            e.cap -= d;
            G[v][e.rev].cap += d;
        }
    }
    return res;
}
void solve()
{
    int s = 0 , t = n - 1;
    V = n;
    for(int i = 0 ; i < m;  i++)
    {
        add_edge(a[i] - 1 , b[i] - 1 , 1  , c[i]);
        add_edge(b[i] - 1 , a[i] - 1 , 1 , c[i]);
    }
    printf("%d\n" , min_cost_flow(s , t , 2));
}
int main()
{
    while(~scanf("%d%d" , &n , &m))
    {
        for(int i = 0  ; i < m ; i++)
        {
            scanf("%d%d%d" , &a[i] , &b[i] , &c[i]);
        }
        solve();
    }
    return 0;
}