uva 1658(最小费用最大流）

题意：一个带权有向图，求起点到终点的两条路径权值之和最小，且两条路径没有公共点（除起点，终点）；

分析：拆点法，将u拆成u和u'，u-u'容量为1，费用为0，这样就能保证每个点只用一次，起点s-s'容量为2，终点t-t'容量为2保证最大流会求出两条路径，若输入u-v,权为c，则增加边u'-v，容量为1，费用为c.

#include <cstdio>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
#define ll long long
#define _cle(m, a) memset(m, a, sizeof(m))
#define repu(i, a, b) for(int i = a; i < b; i++)
#define repd(i, a, b) for(int i = b; i >= a; i--)
#define sfi(n) scanf("%d", &n)
#define pfi(n) printf("%d
", n)
#define sfi2(n, m) scanf("%d%d", &n, &m)
#define pfi2(n, m) printf("%d %d
", n, m)
#define pfi3(a, b, c) printf("%d %d %d
", a, b, c)

#define maxn 2010
#define maxm 10*maxn
const int inf = 0x3f3f3f3f;
struct Nod {
    int b, nxt;
    int cap, cst;
    void init(int b, int nxt, int cap, int cst) {
        this->b = b;
        this->nxt = nxt;
        this->cap = cap;
        this->cst = cst;
    }
};
struct MinCost {
    int E[maxn];        int n;
    Nod buf[maxm*2];    int len;

    int p[maxn];
    void init(int n) {
        this->n = n;
        memset(E, 255, sizeof(E));
        len = 0;
    }
    void addCap(int a, int b, int cap, int cst) {
        buf[len].init(b, E[a], cap, cst);    E[a] = len ++;
        buf[len].init(a, E[b], 0,  -cst);    E[b] = len ++;
    }
    bool spfa(int source, int sink) {
        static queue<int> q;
        static int d[maxn];
        memset(d,  63, sizeof(d));
        memset(p, 255, sizeof(p));

        d[source] = 0;
        q.push(source);
        int u, v;
        while(!q.empty()) {
            u = q.front();
            q.pop();
            for(int i = E[u]; i != -1; i = buf[i].nxt) {
                v = buf[i].b;
                if(buf[i].cap>0 && d[u]+buf[i].cst<d[v]) {
                    d[v] = d[u]+buf[i].cst;
                    p[v] = i;
                    q.push(v);
                }
            }
        }
        return d[sink] != inf;
    }
    int solve(int source, int sink) {
        int minCost = 0,maxFlow = 0;//需要maxFlow的话,想办法返回
        while(spfa(source, sink)) {
            int neck = inf;
            for(int t=p[sink]; t != -1; t = p[ buf[t^1].b ])//buf[t^1].b是父节点
                neck = min(neck, buf[t].cap);
            maxFlow += neck;
            for(int t = p[sink]; t != -1; t = p[ buf[t^1].b ]) {
                buf[t].cap -= neck;
                buf[t^1].cap += neck;
                minCost += buf[t].cst * neck;
            }
        }
        return minCost;
    }
} mc;


int main()
{
    int n, m;
    while(~sfi2(n, m))
    {
        mc.init(n);
        int a, b, c;
        repu(i, 0, m)
        {
            sfi2(a, b), sfi(c);
            mc.addCap((--a) + n, --b, 1, c);
        }
        repu(i, 1, n - 1) mc.addCap(i, i + n, 1, 0);
        mc.addCap(0, n, 2, 0);
        mc.addCap(n - 1, n * 2 - 1, 2, 0);
        pfi(mc.solve(0, n - 1));
    }
    return 0;
}