HDU 3488--Tour【最小费用最大流 && 有向环最小权值覆盖 && 经典】

Tour

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2308    Accepted Submission(s): 1156

Problem Description

In the kingdom of Henryy, there are N (2 <= N <= 200) cities, with M (M <= 30000) one-way roads connecting them. You are lucky enough to have a chance to have a tour in the kingdom. The route should be designed as: The route should contain one or more loops. (A loop is a route like: A->B->……->P->A.)
Every city should be just in one route.
A loop should have at least two cities. In one route, each city should be visited just once. (The only exception is that the first and the last city should be the same and this city is visited twice.)
The total distance the N roads you have chosen should be minimized.

 

Input

An integer T in the first line indicates the number of the test cases.
In each test case, the first line contains two integers N and M, indicating the number of the cities and the one-way roads. Then M lines followed, each line has three integers U, V and W (0 < W <= 10000), indicating that there is a road from U to V, with the distance of W.
It is guaranteed that at least one valid arrangement of the tour is existed.
A blank line is followed after each test case.

 

Output

For each test case, output a line with exactly one integer, which is the minimum total distance.

 

Sample Input

   
   
   
   

    
    
    
    1
6 9
1 2 5
2 3 5
3 1 10
3 4 12
4 1 8
4 6 11
5 4 7
5 6 9
6 5 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    42
   
   
   
   

 

题意：

给出n个点m条单向边边以及经过每条边的费用，让你求出走过一个哈密顿环（除起点外，每个点只能走一次）的最小费用。题目保证至少存在一个环满足条件。

解析：

任意类似的【有向环最小权值覆盖】问题，都可以用最小费用流来写。

由于题目中要求每个点最多走一次，为了防止走多次的发生，我们要把每个点 i 拆成左点i 和 右点i + n两个点。

具体建图如下:

源点outset编号0, 所有左点编号 1~n ，右点编号 n+1 ~ 2*n, 汇点inset 编号 2*n+1.
（1）源点outset到第i个点有边 ( outset, i, 1, 0)，即源点和左点建边。
（2）如果从 i 点到 j 点有权值为 c 的边,那么有边  (i,  j+n,  1,  c)，即左点和右点建边， 确保了每个点只走一次。
（3）每个节点到汇点有边 (i+n,  inset,  1,  0)， 即右点和汇点建边。

最终如果最大流 == n 的话（即满流），那么最小费用就是我们所求，这里题目确保有解，所以下面的代码我就没判断。

为什么这样的构图方法就可以求得我们所要的解， 具体解析请点这里：解析

而且本题时间要求比较严格，各种超时，需要加个去重边处理才行。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define INF 0x3f3f3f3f
#define maxn 500
#define maxm 77000
using namespace std;
int n, m;
int outset;//超级源点
int inset;//超级汇点
struct node {
    int u, v, cap, flow, cost, next;
};

node edge[maxm];
int head[maxn], cnt;
int per[maxn];//记录增广路径上 到达点i的边的编号
int dist[maxn], vis[maxn];

void init(){
    cnt = 0;
    memset(head, -1, sizeof(head));
}

void add(int u, int v, int w, int c){
    int i;
    for(i = head[u]; i != -1; i = edge[i].next){
        node E = edge[i];
        if(v == E.v)
            break;
    }
    if(i != -1){
        if(edge[i].cost > c)
            edge[i].cost = c, edge[i ^ 1].cost = -c;
        return ;
    }
    //edge[cnt] = {u, v, w, 0, c, head[u]}这样写就直接超时了，晕+_+
    node E1 = {u, v, w, 0, c, head[u]};
    edge[cnt] = E1;
    head[u] = cnt++;
    node E2 = {v, u, 0, 0, -c, head[v]};
    edge[cnt] = E2;
    head[v] = cnt++;
}


void getmap(){
    scanf("%d%d", &n, &m);
    outset = 0;
    inset = n * 2 + 1;
    for(int i = 1; i <= n; ++i){
        add(outset, i, 1, 0);
        add(i + n, inset, 1, 0);
    }
    while(m--){
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b + n, 1, c);
    }
}

bool SPFA(int st, int ed){
    queue<int>q;
//    memset(dist, INF, sizeof(dist));
//    memset(vis, 0, sizeof(vis));
//    memset(per, -1, sizeof(per));
    for(int i = 0; i <= inset; ++i){
        dist[i] = INF;
        vis[i] = 0;
        per[i] = -1;
    }
    dist[st] = 0;
    vis[st] = 1;
    q.push(st);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(int i = head[u]; i != -1; i = edge[i].next){
            node E = edge[i];
            if(dist[E.v] > dist[u] + E.cost && E.cap > E.flow){//可以松弛 且 没有满流
                dist[E.v] = dist[u] + E.cost;
                per[E.v] = i;//记录到达这个点的边的编号
                if(!vis[E.v]){
                    vis[E.v] = 1;
                    q.push(E.v);
                }
            }
        }
    }
    return per[ed] != -1;
}

void MCMF(int st, int ed, int &cost, int &flow){
    flow = 0;//总流量
    cost = 0;//总费用
    while(SPFA(st, ed)){//每次寻找花销最小的路径
        int mins = INF;
        for(int i = per[ed]; i != -1; i = per[edge[i ^ 1].v]){
            mins = min(mins, edge[i].cap - edge[i].flow);
        }
         //增广
        for(int i = per[ed]; i != -1; i = per[edge[i ^ 1].v]){
            edge[i].flow += mins;
            edge[i ^ 1].flow -= mins;
            cost += edge[i].cost * mins;
        }
        flow += mins;
    }
}
int main (){
    int T;
    scanf("%d", &T);
    while(T--){
        init();
        getmap();
        int cost, flow;
        MCMF(outset, inset, cost, flow);
        printf("%d\n", cost);
    }
    return 0;
}