两题最小圈覆盖,可以转换成KM来做。
大意:给定一个有向图,把图分成一些环,要求每个点只属于一个环,求满足条件的环所有边权和的最小值。
对于满足条件的环,每个点的入度和出度均为1,我们可以把每个点拆成入点和出点,那么也就是说一个入点对应一个出点,一个出点对应一个入点。那么这个问题就变成了一个最佳匹配问题。
/*HDU 1853*/ #include <iostream> #include <cstdlib> #include <cstdio> #include <string> #include <cstring> #include <cmath> #include <vector> #include <queue> #include <algorithm> #include <map> using namespace std; const int maxn = 110; const int INF = 0x3f3f3f3f; int n, m; int W[maxn][maxn]; int Lx[maxn], Ly[maxn]; int Left[maxn]; bool S[maxn], T[maxn]; bool match(int i) { S[i] = 1; for(int j = 1; j <= m; j++) if(Lx[i]+Ly[j] == W[i][j] && !T[j]) { T[j] = 1; if(!Left[j] || match(Left[j])) { Left[j] = i; return 1; } } return 0; } void update() { int a = INF; for(int i = 1; i <= n; i++) if(S[i]) for(int j = 1; j <= m; 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; } for(int j = 1; j <= m; j++) { if(T[j]) Ly[j] += a; } } void KM() { memset(Left, 0, sizeof(Left)); memset(Lx, 0, sizeof(Lx)); memset(Ly, 0, sizeof(Ly)); for(int i = 1; i <= n; i++) { for(int j = 1; j <= m; j++) Lx[i] = max(Lx[i], W[i][j]); } for(int i = 1; i <= n; i++) { for(;;) { for(int j = 1; j <= m; j++) S[j] = T[j] = 0; if(match(i)) break; else update(); } } } inline void readint(int &x) { char c; c = getchar(); while(!isdigit(c)) c = getchar(); x = 0; while(isdigit(c)) x = x*10+c-'0', c = getchar(); } inline void writeint(int x) { if(x > 9) writeint(x/10); putchar(x%10+'0'); } int N, M; void read_case() { n = m = N; for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) W[i][j] = -INF; for(int i = 1; i <= M; i++) { int u, v, w; readint(u), readint(v), readint(w); W[u][v] = max(W[u][v], -w); } } int cal() { int ans = 0; for(int i = 1; i <= m; i++) { if(Left[i] != 0 && W[Left[i]][i] != -INF) ans -= W[Left[i]][i]; else return -1; } return ans; } void solve() { read_case(); KM(); int ans = cal(); if(ans == -1) printf("-1\n"); else writeint(ans), puts(""); } int main() { while(~scanf("%d%d", &N, &M)) { solve(); } return 0; }