UVa 1658 (拆点法 最小费用流) Admiral

题意：

给出一个有向带权图，求从起点到终点的两条不相交路径使得权值和最小。

分析：

第一次听到“拆点法”这个名词。

把除起点和终点以外的点拆成两个点i和i'，然后在这两点之间连一条容量为1，费用为0的边。这样就保证了每个点最多经过一次。

其他有向边的容量也是1

然后求从起点到终点的流量为2(这样就保证了是两条路径)的最小费用流。

#include <bits/stdc++.h>

using namespace std;

const int maxn = 2000 + 10;
const int INF = 1000000000;

struct Edge
{
int from, to, cap, flow, cost;
Edge(int u, int v, int c, int f, int w): from(u), to(v), cap(c), flow(f), cost(w) {}
};

struct MCMF
{
int n, m;
vector<Edge> edges;
vector<int> G[maxn];
int inq[maxn]; //是否在队列中
int d[maxn];    //Bellman-Ford
int p[maxn];    //上一条弧
int a[maxn];    //可改进量

void Init(int n)
{
this->n = n;
for(int i = 0; i < n; ++i) G[i].clear();
edges.clear();
}

void AddEdge(int from, int to, int cap, int cost)
{
edges.push_back(Edge(from, to, cap, 0, cost));
edges.push_back(Edge(to, from, 0, 0, -cost));
m = edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}

bool BellmanFord(int s, int t, int flow_limit, int& flow, int& cost)
{
for(int i = 0; i < n; ++i) d[i] = INF;
memset(inq, 0, sizeof(inq));
d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;

queue<int> Q;
Q.push(s);
while(!Q.empty())
{
int u = Q.front(); Q.pop();
inq[u] = 0;
for(int i = 0; i < G[u].size(); ++i)
{
Edge& e = edges[G[u][i]];
if(e.cap > e.flow && d[e.to] > d[u] + e.cost)
{
d[e.to] = d[u] + e.cost;
p[e.to] = G[u][i];
a[e.to] = min(a[u], e.cap - e.flow);
if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
}
}
}
if(d[t] == INF) return false;
if(flow + a[t] > flow_limit) a[t] = flow_limit - flow;
flow += a[t];
cost += d[t] * a[t];
for(int u = t; u != s; u = edges[p[u]].from)
{
edges[p[u]].flow += a[t];
edges[p[u]^1].flow -= a[t];
}
return true;
}

int MincostMaxflow(int s, int t, int flow_limit, int& cost)
{
int flow = 0; cost = 0;
while(flow < flow_limit && BellmanFord(s, t, flow_limit, flow, cost));
return flow;
}
}g;

int main()
{
//freopen("in.txt", "r", stdin);

int n, m;
while(scanf("%d%d", &n, &m) == 2 && n)
{
g.Init(n*2-2);
//2~n-1 i和i'的编号分别为1~n-2  n~2n-3
for(int i = 2; i <= n-1; ++i) g.AddEdge(i-1, n-2+i, 1, 0);
for(int i = 0; i < m; ++i)
{   //连接a'->b
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
if(a != 1 && a != n) a += n-2; else a--;
b--;
g.AddEdge(a, b, 1, c);
}
int cost;
g.MincostMaxflow(0, n-1, 2, cost);
printf("%d\n", cost);
}

return 0;
}

代码君