这道题是一个很裸的费用流题目。
一般点之过一次的套路就是拆点连边。
其他的模拟下就水过去了。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define re register 6 #define rep(i, a, b) for (re int i = a; i <= b; ++i) 7 #define repd(i, a, b) for (re int i = a; i >= b; --i) 8 #define maxx(a, b) a = max(a, b); 9 #define minn(a, b) a = min(a, b); 10 #define LL long long 11 #define INF (1 << 30) 12 13 inline int read() { 14 int w = 0, f = 1; char c = getchar(); 15 while (!isdigit(c)) f = c == '-' ? -1 : f, c = getchar(); 16 while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ '0'), c = getchar(); 17 return w * f; 18 } 19 20 const int maxn = 200 + 10, maxm = 20000 + 10; 21 22 struct Edge { 23 int u, v, c, f, w, pre; 24 }; 25 26 int mincost = INF; 27 28 struct Graph { 29 Edge ed[maxm << 2]; 30 int n, m, G[maxn << 1]; 31 void init(int n) { 32 this->n = n; 33 m = 1; 34 memset(G, 0, sizeof(G)); 35 } 36 void add(int u, int v, int c, int w) { 37 ed[++m] = (Edge){u, v, c, 0, w, G[u]}; 38 G[u] = m; 39 ed[++m] = (Edge){v, u, 0, 0, -w, G[v]}; 40 G[v] = m; 41 } 42 int a[maxn << 1], d[maxn << 1], p[maxn << 1], inq[maxn << 1]; 43 int s, t; 44 queue<int> Q; 45 bool spfa(int &flow, int &cost) { 46 memset(inq, 0, sizeof(inq)); 47 rep(i, 1, n) d[i] = INF; 48 d[s] = 0; 49 a[s] = INF; 50 Q.push(s); inq[s] = 1; 51 while (!Q.empty()) { 52 int u = Q.front(); Q.pop(); inq[u] = 0; 53 for (register int i = G[u]; i; i = ed[i].pre) { 54 Edge &e = ed[i]; 55 if (e.f < e.c && d[u] + e.w < d[e.v]) { 56 d[e.v] = d[u] + e.w; 57 a[e.v] = min(a[u], e.c-e.f); 58 p[e.v] = i; 59 if (!inq[e.v]) Q.push(e.v), inq[e.v] = 1; 60 } 61 } 62 } 63 if (d[t] == INF) return false; 64 flow += a[t]; 65 cost += a[t] * d[t]; 66 for (register int i = p[t]; i; i = p[ed[i].u]) { 67 ed[i].f += a[t]; 68 ed[i^1].f -= a[t]; 69 } 70 return true; 71 } 72 int mcmf(int s, int t, int &cost) { 73 this->s = s, this->t = t; 74 int flow = 0; cost = 0; 75 while (spfa(flow, cost)); 76 return flow; 77 } 78 } G; 79 80 int n, m; 81 82 int main() { 83 n = read(), m = read(); G.init(n*2); 84 rep(i, 1, m) { 85 int u = read(), v = read(), w = read(); 86 G.add(u+n, v, 1, w); 87 } 88 rep(i, 2, n-1) G.add(i, i+n, 1, 0); 89 G.add(n, n*2, INF, 0); 90 G.add(1, n+1, INF, 0); 91 int cost; 92 printf("%d", G.mcmf(1, n*2, cost)); 93 printf(" %d", cost); 94 return 0; 95 }