uva 1658 Admiral （最小费最大流）

Posted 2020-10-03 yxysuanfa

uva 1658 Admiral

题目大意：在图中找出两条没有交集的线路，要求这两条线路的费用最小。

解题思路：还是拆点建图的问题。

首先每一个点都要拆成两个点。比如a点拆成a->a’。起点和终点的两点间的容量为2费用为0，保证了仅仅找出两条线路。其余点的容量为1费用为0，保证每点仅仅走一遍，两条线路无交集。然后依据题目给出的要求继续建图。每组数据读入a, b, c， 建立a’到b的边容量为1， 费用为c。图建完之后，用bellman-ford来实现MCMF。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <queue>
using namespace std;
typedef long long ll;
const int N = 2005;
const int INF = 0x3f3f3f3f;
int n, m, s, t;
int a[N], pre[N], d[N], inq[N]; 

struct Edge{
    int from, to, cap, flow;
    ll cos;
};

vector<Edge> edges;
vector<int> G[N];

void init() {
    for (int i = 0; i < 2 * n; i++) G[i].clear();
    edges.clear();
}

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

void input() {
    addEdge(1, n + 1, 2, 0, 0);
    for (int i = 2; i <= n - 1; i++) {
        addEdge(i, i + n, 1, 0, 0); 
    }
    addEdge(n, 2 * n, 2, 0, 0);
    int u, v;
    ll c;
    for (int i = 0; i < m; i++) {
        scanf("%d %d %lld", &u, &v, &c);    
        addEdge(u + n, v, 1, 0, c);
    }
}

int BF(int s, int t, int& flow, ll& cost) {
    queue<int> Q;
    memset(inq, 0, sizeof(inq));
    memset(a, 0, sizeof(a));
    memset(pre, 0, sizeof(pre));
    for (int i = 0; i <= 2 * n + 1; i++) d[i] = INF;
    d[s] = 0;
    a[s] = INF;
    inq[s] = 1;
    int flag = 1;
    pre[s] = 0;
    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.cos) {
                d[e.to] = d[u] + e.cos;
                a[e.to] = min(a[u], e.cap - e.flow);
                pre[e.to] = G[u][i];
                if (!inq[e.to]) {
                    inq[e.to] = 1;
                    Q.push(e.to);
                }
            }   
        }
        flag = 0;
    }
    if (d[t] == INF) return 0;
    flow += a[t];
    cost += (ll)d[t] * (ll)a[t];
    for (int u = t; u != s; u = edges[pre[u]].from) {
        edges[pre[u]].flow += a[t];
        edges[pre[u]^1].flow -= a[t];
    }
    return 1;
}

int MCMF(int s, int t, ll& cost) {
    int flow = 0;
    cost = 0;       
    while (BF(s, t, flow, cost));
    return flow;
}

int main() {
    while (scanf("%d %d", &n, &m) == 2) {
        s = 1, t = 2 * n;   
        ll cost;
        init();
        input();
        MCMF(s, t, cost);
        printf("%lld\n", cost);
    }
    return 0;
}