【luogu P3381 最小费用最大流】 模板

题目链接：https://www.luogu.org/problemnew/show/P3381

把bfs变成spfa

 1 #include <queue>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <algorithm>
 6 using namespace std;
 7 const int maxn = 1000000 + 10;
 8 int n, m, s, t, maxflow, mincost, dis[maxn], flow[maxn], pre[maxn], last[maxn];
 9 struct edge{
10     int to, next, cost, flow;
11 }e[maxn<<2];
12 int head[maxn], cnt = -1;
13 bool vis[maxn];
14 queue<int> q;
15 void add(int u, int v, int w, int c)
16 {
17     e[++cnt].next = head[u];
18     e[cnt].to = v;
19     e[cnt].cost = c;
20     e[cnt].flow = w;
21     head[u] = cnt;
22     
23     e[++cnt].next = head[v];
24     e[cnt].to = u;
25     e[cnt].cost = -c;
26     e[cnt].flow = 0;
27     head[v] = cnt;
28 }
29 bool SPFA(int s, int t)
30 {
31     memset(dis, 0x7f, sizeof(dis));
32     memset(flow, 0x7f, sizeof(flow));
33     memset(vis, 0, sizeof(vis));
34     q.push(s); pre[t] = -1; vis[s] = 1; dis[s] = 0;
35     while(!q.empty())
36     {
37         int now = q.front(); q.pop();
38         vis[now] = 0;
39         for(int i = head[now]; i != -1; i = e[i].next)
40         {
41             if(e[i].flow > 0 && dis[e[i].to] > dis[now] + e[i].cost)
42             {
43                 dis[e[i].to] = dis[now] + e[i].cost;
44                 pre[e[i].to] = now;
45                 last[e[i].to] = i;
46                 flow[e[i].to] = min(flow[now], e[i].flow);
47                 if(!vis[e[i].to])
48                 {
49                     q.push(e[i].to);
50                     vis[e[i].to] = 1;
51                 }
52             }
53         }
54     }
55     return pre[t] != -1;
56 }
57 void MCMF(int s, int t)
58 {
59     while(SPFA(s, t))
60     {
61         int now = t;
62         maxflow += flow[t];
63         mincost += flow[t] * dis[t];
64         while(now != s)
65         {
66             e[last[now]].flow -= flow[t];
67             e[last[now]^1].flow += flow[t];
68             now = pre[now];
69         }
70     }
71 }
72 int main()
73 {
74     memset(head, -1, sizeof(head));
75     scanf("%d%d%d%d",&n,&m,&s,&t);
76     for(int i = 1; i <= m; i++)
77     {
78         int u, v, w, c;
79         scanf("%d%d%d%d",&u,&v,&w,&c);
80         add(u, v, w, c);
81     }
82     MCMF(s, t);
83     printf("%d %d\n",maxflow, mincost);
84     return 0;
85 }

 

posted @ 2018-05-16 19:55  Misaka_Azusa  阅读(171)  评论(0)  编辑  收藏  举报
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