ZOJ 2342 Roads 二分图匹配

来源：互联网发布：全职法师实体书淘宝编辑：程序博客网时间：2024/06/06 06:34

题意：给你一个n个点m条边的带权无向图其前n-1条边构成一个生成树问你修改各条边的值使得前n-1条边构成最小生成树并且各条边的改变量之和最小

思路：既然是要求前n-1条边组成最小生成树则这前n-1条边权值一定不能增大而剩余的边权值一定不能减少对于前n-1条边权值w[a]和剩下的边w[b] 使得 w[a]-d[a]<w[b]+d[b] 移项得 w[a]-w[b]<d[a]+d[b] 着看起来是不是有点像KM算法里的可行顶标这里也是本题的精髓对于前n-1条边建立一棵生成树后每加一条边就会产生一条回路为了使得新加入的边的权值在所在的圈中最大就要对圈中所有w[a]>w[b]的边改变其权值w[a]-d[a]<w[b]+d[b] 最后直接套KM即可

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;const int maxn = 400 + 10;int cas = 0;struct Edge{          int v, id, d, next;          Edge(int v, int id, int d, int next) : v(v), id(id), d(d), next(next) {}          Edge() {}};struct WORK{          int w[maxn][maxn], n;          int Lx[maxn], Ly[maxn];          int left[maxn];          int cntE, Head[maxn];          int c[maxn*2];          bool S[maxn], T[maxn], vis[maxn];          int N, M;          Edge e[maxn*2];          void init(int n){                    this -> n = n;                    cntE = 0;                    memset(Head, -1, sizeof(Head));          }          void Add(int u, int v, int id, int d){                    e[cntE] = Edge(v, id, d, Head[u]);                    Head[u] = cntE++;                    e[cntE] = Edge(u, id, d, Head[v]);                    Head[v] = cntE++;          }          bool match(int i){                    S[i] = true;                    for(int j = 1; j <= n; j++)if(Lx[i] + Ly[j] == w[i][j] && !T[j]){                              T[j] = true;                              if(!left[j] || match(left[j])){                                        left[j] = i;                                        return true;                              }                    }                    return false;          }          void update(){                    int a = 1 << 30;                    for(int i = 1; i <= n; i++)if(S[i])                              for(int j = 1; j <= n; j++)if(!T[j])                              a = min(a, Lx[i] + Ly[j] - w[i][j]);                    for(int i = 1; i <= n; i++){                              if(S[i]) Lx[i] -= a;                              if(T[i]) Ly[i] += a;                    }          }          void KM(){                    for(int i = 1; i <= n; i++){                              left[i] = Lx[i] = Ly[i] = 0;                              for(int j = 1; j <= n; j++) Lx[i] = max(Lx[i], w[i][j]);                    }                    for(int i = 1; i <= n; i++){                              for(;;){                                        for(int j = 1; j <= n; j++) S[j] = T[j] = 0;                                        if(match(i)) break;                                        else update();                              }                    }          }          bool build(int s, int t, int id, int d){                    if(s == t) return true;                    vis[s] = true;                    for(int i = Head[s]; ~i; i = e[i].next){                              if(vis[e[i].v]) continue;                              if(build(e[i].v, t, id, d)){                                        if(e[i].d > d) w[e[i].id][id] = e[i].d - d;                                        return true;                              }                    }                    return false;          }          void solve(){                    scanf("%d%d", &N, &M);                    init( max(N - 1, M - N + 1) );                    memset(w, 0, sizeof(w));                    int u, v, d;                    for(int i = 1; i < N; i++){                              scanf("%d%d%d", &u, &v, &d);                              c[i] = d;                              Add(u, v, i, d);                    }                    for(int i = N; i <= M; i++){                              scanf("%d%d%d", &u, &v, &d);                              c[i] = d;                              memset(vis, false, sizeof(vis));                              build(u, v, i - N + 1, d);                    }                    KM();                    output();          }          void output(){                    if(cas++) printf("\n");                    for(int i = 1; i <= M; i++){                              if(i < N) printf("%d\n", c[i] - Lx[i]);                              else printf("%d\n", c[i] + Ly[i-N+1]);                    }          }}solver;int main(){          //freopen("in.txt", "r", stdin);          int T;          scanf("%d", &T);          while(T--) solver.solve();          return 0;}

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