BZOJ3396 [Usaco2009 Jan]Total flow 水流

虽然我想说，这貌似是。。。可以直接dfs做的。。。

但是还是Dinic最大流保险一点。。。

板子补完中→_→

 

  1 /**************************************************************
  2     Problem: 3396
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:0 ms
  7     Memory:824 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <cctype>
 12 #include <cstring>
 13 #include <algorithm>
 14  
 15 using namespace std;
 16 const int inf = (int) 1e9;
 17 const int N = 65;
 18 const int M = 1505;
 19  
 20 struct edges{
 21     int next, to, f;
 22     edges() {}
 23     edges(int _next, int _to, int _f) : next(_next), to(_to), f(_f) {}
 24 } e[M];
 25  
 26 int n;
 27 int first[N], tot = 1, d[N], q[N];
 28 int S, T;
 29  
 30 inline int read() {
 31     int x = 0;
 32     char ch = getchar();
 33     while (ch < '0' || '9' < ch)
 34         ch = getchar();
 35     while ('0' <= ch && ch <= '9') {
 36         x = x * 10 + ch - '0';
 37         ch = getchar();
 38     }
 39     return x;
 40 }
 41  
 42 inline char getc() {
 43     char ch = getchar();
 44     while (!isalpha(ch))
 45         ch = getchar();
 46     return ch;
 47 }
 48  
 49 inline void add_edge(int x, int y, int z){
 50     e[++tot] = edges(first[x], y, z);
 51     first[x] = tot;
 52 }
 53    
 54 inline void Add_Edges(int x, int y, int z){
 55     add_edge(x, y, z);
 56     add_edge(y, x, 0);
 57 }
 58  
 59 bool bfs(){
 60     memset(d, 0, sizeof(d));
 61     q[1] = S, d[S] = 1;
 62     int l = 0, r = 1, x, y;
 63     while (l < r){
 64         ++l;
 65         for (x = first[q[l]]; x; x = e[x].next){
 66             y = e[x].to;
 67             if (!d[y] && e[x].f)
 68                 q[++r] = y, d[y] = d[q[l]] + 1;
 69         }
 70     }
 71     return d[T];
 72 }
 73    
 74 int dinic(int p, int limit){
 75     if (p == T || !limit) return limit;
 76     int x, y, tmp, rest = limit;
 77     for (x = first[p]; x; x = e[x].next){
 78         y = e[x].to;
 79         if (d[y] == d[p] + 1 && e[x].f && rest){
 80             tmp = dinic(y, min(rest, e[x].f));
 81             rest -= tmp;
 82             e[x].f -= tmp, e[x ^ 1].f += tmp;
 83             if (!rest) return limit;
 84         }
 85     }
 86     if (limit == rest) d[p] = 0;
 87     return limit - rest;
 88 }
 89    
 90 int Dinic(){
 91     int res = 0, x;
 92     while (bfs())
 93         res += dinic(S, inf);
 94     return res;
 95 }
 96  
 97 int main() {
 98     char ch;
 99     int i, x, y, z;
100     n = read();
101     for (i = 1; i <= n; ++i) {
102         ch = getc();
103         if ('A' <= ch && ch <= 'Z')
104             x = ch - 'A' + 1;
105         else x = ch - 'a' + 27;
106         ch = getc();
107         if ('A' <= ch && ch <= 'Z')
108             y = ch - 'A' + 1;
109         else y = ch - 'a' + 27;
110         z = read();
111         Add_Edges(x, y, z);
112     }
113     S = 1, T = 26;
114     printf("%d\n", Dinic());
115     return 0;
116 }
View Code

（p.s. 为毛线。。。之前的Dinic写错了还过了？233） 

posted on 2014-11-16 21:07  Xs酱~  阅读(224)  评论(0)  编辑  收藏  举报