uva 10594 - Data Flow(费用流）

来源：互联网发布：傅园慧网络直播回看编辑：程序博客网时间：2024/04/29 22:26

题目链接：uva 10594 - Data Flow

题目大意：给出一张图，然后给出K和D，D表示每条路的容量，然后由点1到点n运送数据，流量为K时的最小费用。

解题思路：和uva 10806 - Dijkstra, Dijkstra.做法一个样.....表示说不解释，就在起点0和1之间加条容量为K的路就好。

#include <stdio.h>#include <string.h>#include <queue>#define ll long long#define min(a,b) (a)<(b)?(a):(b)using  namespace std;const int MAXN = 50000;ll INF = 1000000000000000;const int N = 105;struct edge {int next, im;int far, son;ll cap, flow;ll cost;}s[MAXN];int n, m, tmp, h[N];ll K, D;void add(int x, int y, ll c, ll cost, int im) {s[tmp].next = h[x];h[x] = tmp;s[tmp].im = tmp + im;s[tmp].far = x;s[tmp].son = y;s[tmp].cap = c;s[tmp].flow = 0;s[tmp].cost = cost;tmp++;}void init() {ll a[MAXN], b[MAXN], c[MAXN];tmp = 0;memset(h, -1, sizeof(h));for (int i = 0; i < m; i++)scanf("%lld%lld%lld", &a[i], &b[i], &c[i]);scanf("%lld%lld", &K, &D);add(0, 1, K, 0, 1);add(1, 0, 0, 0, -1);for (int i = 0; i < m; i++) {add(a[i], b[i], D, c[i], 1);add(b[i], a[i], 0, -c[i], -1);add(b[i], a[i], D, c[i], 1);add(a[i], b[i], 0, -c[i], -1);}}void solve() {queue<int> que;ll ans = 0, f = 0, d[N];int u, vis[N], path[MAXN];while (1) {for (int i = 0; i <= n + 1; i++) d[i] = (i == 0) ? 0 : INF;memset(path, -1, sizeof(path));memset(vis, 0, sizeof(vis));que.push(0);vis[0] = 1;while ( !que.empty() ) {u = que.front(), que.pop();vis[u] = 0;for (int i = h[u]; i != -1; i = s[i].next) {int v = s[i].son;if ( s[i].cap > s[i].flow && d[v] > d[u] + s[i].cost) {d[v] = d[u] + s[i].cost;path[v] = i;if (vis[v] == 0) {que.push(v);vis[v] = 1;}}}}if (d[n] == INF) break;ll a = INF;for (int i = n; i; i = s[path[i]].far)a = min(a, s[path[i]].cap - s[path[i]].flow);f += a;ans += a * d[n];for (int i = n; i; i = s[path[i]].far) {s[path[i]].flow += a;int im = s[path[i]].im;s[im].flow -= a;}}if (f < K)printf("Impossible.\n");elseprintf("%lld\n", ans);}int main () {while (scanf("%d%d", &n, &m) == 2) {init();solve();}return 0;}