「网络流 24 题」餐巾计划

这道题目很关键的一点就是，要明白用的餐巾和洗的餐巾是两个系统。

所以我们把每一天所用餐巾直接用一个源点和汇点连接。解决第一个系统。

此时可能会洗餐巾，所以我们另起一个点，从源点每一天引出脏的餐巾，数目依然是当天所用的餐巾数量，然后从这个点连接到m天后的第一系统的点和n天后第一系统的点，费用分别为f和s，流量为无穷大。当天的餐巾就算不洗也可以留下来，所以我们还需要把这些餐巾连接起来，当天不洗的餐巾连一条线到下一天不洗的餐巾处，流量为无穷大，费用为0.第二个系统就是辅助第一个系统的。

代码如下：

#include<iostream>#include<cstdio>#include<vector>#include<queue>#include<utility>#include<stack>#include<algorithm>#include<cstring>#include<string>#include<cmath>#include<set>#include<map>using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 2500 + 5;const int maxm = 10000 + 5;int N;int head[maxn], to[maxm], front[maxm], flow[maxm], cost[maxm], ppp;int dis[maxn], minflow[maxn];bool flag[maxn];pair<int, int> par[maxn];struct MIN_COST_MAX_FLOW{void init() {memset(head, -1, sizeof(head));ppp = 0;}bool spfa(int s, int t) {int u, v;fill(dis, dis + t + 1, INF);memset(flag, 0, sizeof(flag));dis[s] = 0;minflow[s] = INF;queue <int> q;q.push(s);while(!q.empty()) {u = q.front();q.pop();flag[u] = 0;for(int i = head[u]; ~i; i = front[i]) {v = to[i];if(flow[i] && dis[v] > dis[u] + cost[i]) {dis[v] = dis[u] + cost[i];par[v] = (make_pair(u, i));minflow[v] = min(minflow[u], flow[i]);if(!flag[v]) {flag[v] = 1;q.push(v);}}}}if(dis[t] == INF)return 0;return 1;}int slove(int s, int t) {int ans = 0, p;while(spfa(s, t)) {p = t;while(p != s) {flow[par[p].second] -= minflow[t];flow[par[p].second^1] += minflow[t];p = par[p].first;}ans += dis[t] * minflow[t];}return ans;}void add_edge(int u, int v, int f, int c) {to[ppp] = v, front[ppp] = head[u], flow[ppp] = f, cost[ppp] = c, head[u] = ppp++;to[ppp] = u, front[ppp] = head[v], flow[ppp] = 0, cost[ppp] = -c, head[v] = ppp++;}}mcmf;int main() {#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);//    freopen("out.txt", "w", stdout);#endifint p, m, f, n, s;mcmf.init();scanf("%d%d%d%d%d%d", &N, &p, &m, &f, &n, &s);int start = 2 * N, tank = start + 1;for(int i = 0; i < N; i++) {int tmp, a = i * 2, b = a + 1;scanf("%d", &tmp);mcmf.add_edge(start, a, tmp, p);mcmf.add_edge(a, b, tmp, 0);mcmf.add_edge(b, tank, tmp, 0);if(i + m < N)mcmf.add_edge(b, (i + m) * 2, INF, f);if(i + n < N)mcmf.add_edge(b, (i + n) * 2, INF, s);if(i + 1 < N)mcmf.add_edge(a, a + 2, INF, 0);}int ans = mcmf.slove(start, tank);printf("%d\n", ans);return 0;}