• poj 2135 Farm Tour 费用流


    题目链接

    给一个图, N个点, m条边, 每条边有权值, 从1走到n, 然后从n走到1, 一条路不能走两次,求最短路径。

    如果(u, v)之间有边, 那么加边(u, v, 1, val), (v, u, 1, val), val是路的长度,代表费用, 1是流量。

      1 #include <iostream>
      2 #include <vector>
      3 #include <cstdio>
      4 #include <cstring>
      5 #include <algorithm>
      6 #include <cmath>
      7 #include <map>
      8 #include <set>
      9 #include <string>
     10 #include <queue>
     11 using namespace std;
     12 #define pb(x) push_back(x)
     13 #define ll long long
     14 #define mk(x, y) make_pair(x, y)
     15 #define lson l, m, rt<<1
     16 #define mem(a) memset(a, 0, sizeof(a))
     17 #define rson m+1, r, rt<<1|1
     18 #define mem1(a) memset(a, -1, sizeof(a))
     19 #define mem2(a) memset(a, 0x3f, sizeof(a))
     20 #define rep(i, a, n) for(int i = a; i<n; i++)
     21 #define ull unsigned long long
     22 typedef pair<int, int> pll;
     23 const double PI = acos(-1.0);
     24 const double eps = 1e-8;
     25 const int mod = 1e9+7;
     26 const int inf = 1061109567;
     27 const int dir[][2] = { {1, 0}, {0, 1}, {0, -1}, {0, 1} };
     28 const int maxn = 4e5+5;
     29 int num, head[maxn*2], s, t, n, m, nn, dis[maxn], flow, cost, cnt, cap[maxn], q[maxn], cur[maxn], vis[maxn];
     30 struct node
     31 {
     32     int to, nextt, c, w;
     33     node(){}
     34     node(int to, int nextt, int c, int w):to(to), nextt(nextt), c(c), w(w) {}
     35 }e[maxn*2];
     36 int spfa() {
     37     int st, ed;
     38     st = ed = 0;
     39     mem2(dis);
     40     ++cnt;
     41     dis[s] = 0;
     42     cap[s] = inf;
     43     cur[s] = -1;
     44     q[ed++] = s;
     45     while(st<ed) {
     46         int u = q[st++];
     47         vis[u] = cnt-1;
     48         for(int i = head[u]; ~i; i = e[i].nextt) {
     49             int v = e[i].to, c = e[i].c, w = e[i].w;
     50             if(c && dis[v]>dis[u]+w) {
     51                 dis[v] = dis[u]+w;
     52                 cap[v] = min(c, cap[u]);
     53                 cur[v] = i;
     54                 if(vis[v] != cnt) {
     55                     vis[v] = cnt;
     56                     q[ed++] = v;
     57                 }
     58             }
     59         }
     60     }
     61     if(dis[t] == inf)
     62         return 0;
     63     cost += dis[t]*cap[t];
     64     flow += cap[t];
     65     for(int i = cur[t]; ~i; i = cur[e[i^1].to]) {
     66         e[i].c -= cap[t];
     67         e[i^1].c += cap[t];
     68     }
     69     return 1;
     70 }
     71 int mcmf() {
     72     flow = cost = 0;
     73     while(spfa())
     74         ;
     75     return cost;
     76 }
     77 void add(int u, int v, int c, int val) {
     78     e[num] = node(v, head[u], c, val); head[u] = num++;
     79     e[num] = node(u, head[v], 0, -val); head[v] = num++;
     80 }
     81 void init() {
     82     mem1(head);
     83     num = cnt = 0;
     84     mem(vis);
     85 }
     86 void input() {
     87     int x, y, w;
     88     while(m--) {
     89         scanf("%d%d%d", &x, &y, &w);
     90         add(x, y, 1, w);
     91         add(y, x, 1, w);
     92     }
     93     add(s, 1, 2, 0);
     94     add(n, t, 2, 0);
     95 }
     96 int main()
     97 {
     98     while(~scanf("%d%d", &n, &m)) {
     99         init();
    100         s = 0, t = n+1;
    101         input();
    102         int ans = mcmf();
    103         printf("%d
    ", ans);
    104     }
    105     return 0;
    106 }
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  • 原文地址:https://www.cnblogs.com/yohaha/p/5069598.html
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