HDU 1853 - Cyclic Tour (最小费用最大流)

题意

求把每个点都属于一个环,环的最小权和。

思路

学习了最小费用最大流求环的最小权的方法。

把一个点拆成两个,控制入度和出度,如果最后的流量等于一开始的流量,就可以。

不过奇怪的是在另外一题完全一样的题里我的模板竟然TLE了。。。再多做几题看看

代码

#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
#define LL long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 300 + 10;
const int MOD = 1e9 + 7;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
 
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
 
struct EDGE
{
    int from, to, cap, flow, cost;
};
 
struct MCMF
{
    int m, st, ed;
    vector<EDGE> edges;
    int head[MAXN], next[MAXN * 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->ed = n;
        MS(head, -1);
        edges.clear();
    }
 
    void add_edge(int from, int to, int cap, int cost)
    {
        edges.PB((EDGE){from, to, cap, 0, cost});
        edges.PB((EDGE){to, from, 0, 0, -cost});
        m = SZ(edges);
        next[m - 2] = head[from]; head[from] = m - 2;
        next[m - 1] = head[to]; head[to] = m - 1;
    }
 
    bool SPFA(int st, int ed, int &flow, int &cost)
    {
        fill(d, d + ed + 1, INF);
        MS(inq, 0);
        d[st] = 0; inq[st] = 1; p[st] = -1; a[st] = INF;
        queue<int> qu;
        qu.push(st);
        while (!qu.empty())
        {
            int u = qu.front(); qu.pop();
            inq[u] = 0;
            for (int i = head[u]; i != -1; i = next[i])
            {
                EDGE &e = edges[i];
                if (e.cap > e.flow && d[e.to] > d[u] + e.cost)
                {
                    d[e.to] = d[u] + e.cost;
                    p[e.to] = i;
                    a[e.to] = min(a[u], e.cap - e.flow);
                    if (!inq[e.to])
                        qu.push(e.to), inq[e.to] = 1;
                }
            }
        }
        if (d[ed] == INF) return false;
        flow += a[ed]; cost += d[ed] * a[ed];
        int u = ed;
        while (u != st)
        {
            edges[p[u]].flow += a[ed];
            edges[p[u] ^ 1].flow -= a[ed];
            u = edges[p[u]].from;
        }
        return true;
    }
 
}MFMC;
 
int main()
{
    //ROP;
    int T, i, j, ncity, nway;
    while (~scanf("%d%d", &ncity, &nway))
    {
        int st = 0, ed = 2 * ncity + 1;
        MFMC.init(ed);
        while (nway--)
        {
            int a, b, c;
            scanf("%d%d%d", &a, &b, &c);
            MFMC.add_edge(a, b + ncity, 1, c);
        }
        for (i = 1; i <= ncity; i++)
        {
            MFMC.add_edge(st, i, 1, 0);
            MFMC.add_edge(i + ncity, ed, 1, 0);
            //MFMC.add_edge(i, i + ncity, 1, 0);
        }
        int cost = 0, flow = 0;
        while (MFMC.SPFA(st, ed, flow, cost));
        if (flow == ncity) printf("%d\n", cost);
        else printf("-1\n");
    }
    return 0;
}

Powered by Jekyll and Theme by solid