HDU 4411 Arrest 费用流

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题目链接：点击打开链接

题意:

给定n+1个点（[0,n] )m条边的无向图。起点为0，k个人初始在起点，

去遍历图使得每个点至少被一人走过且遍历 i 点时 i-1 必须已经被遍历。

使得k人的路径和最小，最后k人要回到起点。

思路：

费用流，因为对于一个人来说，这个人遍历点的序列一定是一个递增序列（不需要连续）

所以建图时i的出点只需要连接i+? 的入点。

若建一个完全图则会因为spfa跑负环。。。

#include<iostream>#include<stdio.h>#include<string.h>#include<queue>#include<math.h>using namespace std;#define ll int#define inf 1000000#define Inf 0x3FFFFFFFFFFFFFFFLL  #define N 250#define M N*N*4struct Edge {ll to, cap, cost, nex;Edge(){}Edge(ll to, ll cap, ll cost, ll next) :to(to), cap(cap), cost(cost), nex(next){}} edge[M << 1];ll head[N], edgenum;ll D[N], A[N], P[N];bool inq[N];void add(ll from, ll to, ll cap, ll cost) {edge[edgenum] = Edge(to, cap, cost, head[from]);head[from] = edgenum++;edge[edgenum] = Edge(from, 0, -cost, head[to]);head[to] = edgenum++;}bool spfa(ll s, ll t, ll &flow, ll &cost) {for (ll i = 0; i <= t; i++) D[i] = inf;memset(inq, 0, sizeof inq);queue<ll>q;q.push(s);D[s] = 0; A[s] = inf;while (!q.empty()) {ll u = q.front(); q.pop();inq[u] = 0;for (ll i = head[u]; ~i; i = edge[i].nex){Edge &e = edge[i];if (e.cap && D[e.to] > D[u] + e.cost){D[e.to] = D[u] + e.cost;P[e.to] = i;A[e.to] = min(A[u], e.cap);if (!inq[e.to]){inq[e.to] = 1; q.push(e.to);}}}}//若费用为inf则中止费用流if (D[t] == inf) return false;cost += D[t] * A[t];flow += A[t];ll u = t;while (u != s) {edge[P[u]].cap -= A[t];edge[P[u] ^ 1].cap += A[t];u = edge[P[u] ^ 1].to;}return true;}ll Mincost(ll s, ll t){ll flow = 0, cost = 0;while (spfa(s, t, flow, cost));return cost;}void init(){ memset(head, -1, sizeof head); edgenum = 0; }const int MAXN = 105;int g[MAXN][MAXN];int n, m, k;void floyd() {for (int k = 0; k <= n; k++) {for (int i = 0; i <= n; i++) {for (int j = 0; j <= n; j++) {if (g[i][j] > g[i][k] + g[k][j]) {g[i][j] = g[i][k] + g[k][j];}}}}}int main(){while (scanf("%d%d%d", &n, &m, &k) != EOF) {if (n == 0 && m == 0 && k == 0) break;for (int i = 0; i <= n; i++){for (int j = 0; j <= n; j++)g[i][j] = 10000005;g[i][i] = 0;}for (int i = 0, u, v, w; i < m; i++) {scanf("%d%d%d", &u, &v, &w);g[u][v] = g[v][u] = min(w, g[u][v]);}floyd();init();int to = n * 2 + 3, from = n * 2 + 2;for (int i = 1; i <= n; i++){add(0, i * 2, 1, g[0][i]);add(i * 2, i * 2 + 1, 1, -inf);add(i * 2 + 1, to, 1, g[i][0]);for (int j = i + 1; j <= n; j++)add(i * 2 + 1, j * 2, 1, g[i][j]);}add(from, 0, k, 0);add(0, to, k, 0);printf("%d\n", Mincost(from, to) + inf*n);}return 0;}

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