hdu 3488（uva 1349）（KM）

这道题是uva 1349 的简化版，那题没过，不知道为什么。我觉得那题就是多了一个先判断他最大匹配数是不是n，是的话，再找最优匹配。

回到这题，匹配问题，又是有向图，直接想到了拆点法。然后发现若每个点都恰好属于一个环，即是每个点的初度入度都为1，所以只要这个二分图有完美匹配，就是满足题目意思的。

这样的话就是找最小匹配了，处理方法，每条边的距离用一个INF值减，最后累加时别忘了再用INF减回来。

Tour

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1156 Accepted Submission(s): 608

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
   
   
   
   

Source

2010 ACM-ICPC Multi-University Training Contest（6）——Host by BIT

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zhouzeyong

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int maxn = 200 + 5; // 顶点的最大数目
const int INF = 1000000000;

// 最大权匹配
struct KM {
  int n;                  // 左右顶点个数
  vector<int> G[maxn];    // 邻接表
  int W[maxn][maxn];      // 权值
  int Lx[maxn], Ly[maxn]; // 顶标
  int left[maxn];         // left[i]为右边第i个点的匹配点编号，-1表示不存在
  bool S[maxn], T[maxn];  // S[i]和T[i]为左/右第i个点是否已标记

  void init(int n) {
    this->n = n;
    for(int i = 0; i < n; i++) G[i].clear();
    memset(W, 0, sizeof(W));
  }

  void AddEdge(int u, int v, int w) {
    G[u].push_back(v);
    W[u][v] = w;
  }

  bool match(int u){
    S[u] = true;
    for(int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      if (Lx[u]+Ly[v] == W[u][v] && !T[v]){
        T[v] = true;
        if (left[v] == -1 || match(left[v])){
          left[v] = u;
          return true;
        }
      }
    }
    return false;
  }

  void update(){
    int a = INF;
    for(int u = 0; u < n; u++) if(S[u])
      for(int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if(!T[v]) a = min(a, Lx[u]+Ly[v] - W[u][v]);
      }
    for(int i = 0; i < n; i++) {
      if(S[i]) Lx[i] -= a;
      if(T[i]) Ly[i] += a;
    }
  }

  void solve() {
    for(int i = 0; i < n; i++) {
      Lx[i] = *max_element(W[i], W[i]+n);
      left[i] = -1;
      Ly[i] = 0;
    }
    for(int u = 0; u < n; u++) {
      for(;;) {
        for(int i = 0; i < n; i++) S[i] = T[i] = false;
        if(match(u)) break;else update();
      }
    }printf("END\n");
  }
};

KM solver;

int main(){
    int t,n,m;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        solver.init(n);
        while(m--){
            int from,to,dist;
            scanf("%d%d%d",&from,&to,&dist);from--;to--;
            if(solver.W[from][to] != 0){
                dist = max(INF-dist,solver.W[from][to]);
            }
            else
                dist = INF - dist;
            solver.AddEdge(from,to,dist);
        }
        solver.solve();
        int ans = 0;
        for(int i = 0;i < n;i++)
            ans += INF-solver.W[solver.left[i]][i];
        printf("%d\n",ans);
    }
    return 0;
}