[SDOI2009]晨跑

来源：互联网发布：淘宝销量代码编辑：程序博客网时间：2024/05/16 15:35

看懂题意就很简单了233~~以前语文差的看不懂题意~~

将每个点拆成出点入点,流量为1费用为0,控制每个点经过一次,然后最小费用最大流

Code:

#include <cstdio>#include <queue>#include <cstring>#include <bitset>using namespace std;#define M 100001#define N 10001#define INF (0x7FFFFFFF)#define S(x) (x<<1)#define T(x) ((x<<1)-1)struct E{int to, f, c, next, op;}G[M];int point[N], cnt, S, T, f, n, m, cost, D[N], path[N], prev[N];bitset<N> vis;queue<int> Q;inline void add(int u, int v, int f, int c){cnt++, G[cnt] = (E){v, f, c, point[u], cnt+1}, point[u] = cnt;cnt++, G[cnt] = (E){u, 0, -c, point[v], cnt-1}, point[v] = cnt;}inline bool spfa(){for (int i = S;i <= T; ++i)D[i] = INF;D[S] = 0;vis[S] = 1;prev[S] = -1;Q.push(S);while (!Q.empty()){int u = Q.front();Q.pop();for (int i = point[u];i;i = G[i].next){int v = G[i].to;if (G[i].f > 0 && D[v] > D[u] + G[i].c){D[v] = D[u] + G[i].c;path[v] = i;prev[v] = u;if (!vis[v])vis[v] = 1, Q.push(v);}}vis[u] = 0;}return D[T] != INF;}inline int argument(){int delta(INF), i, j, res(0);for (i = T;prev[i] != -1;i = prev[i]){j = path[i];delta = min(delta, G[j].f);}f += delta;for (i = T;prev[i] != -1;i = prev[i]){j = path[i];G[j].f -= delta;G[G[j].op].f += delta;res += delta * G[j].c;}return res;}int main(){scanf("%d %d", &n, &m);S = T(1), T = S(n);for (int i = 1;i <= m; ++i){int a, b, c, x(1), y(n);scanf("%d %d %d", &a, &b, &c);x = T(a), y = S(b);add(x, y, 1, c);}for (int i = 1;i <= n; ++i)if (i != 1 && i != n)add(S(i), T(i), 1, 0);while (spfa())cost += argument();printf("%d %d\n", f, cost);}

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