BZOJ1001 [BeiJing2006]狼抓兔子

一眼最小割，转化成最大流来做。

然后发现点数达到10^6级别，妥妥TLE，于是需要进一步思考。

由网上大量题解可知，一个图的最大流等于它的对偶图的最短路，于是只要Dijkstra就可以了。

建图有点恶心。。。查了好长时间。。。

 

  1 /**************************************************************
  2     Problem: 1001
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:520 ms
  7     Memory:94588 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <cstring>
 12 #include <algorithm>
 13 #include <queue>
 14  
 15 using namespace std;
 16  
 17 struct edges{
 18     int next, to, v;
 19 } e[6000005];
 20  
 21 struct heap_node{
 22     int v, to;
 23 };
 24 inline bool operator < (const heap_node &a, const heap_node &b){
 25     return a.v > b.v;
 26 }
 27  
 28 priority_queue <heap_node> h;
 29 heap_node NODE;
 30 int tot, S, T, n, m;
 31 int d[3000005], first[3000005];
 32 char ch;
 33  
 34 inline int read(){
 35     int x = 0, sgn = 1;
 36     ch = getchar();
 37     while (ch < '0' || ch > '9'){
 38         if (ch == '-') sgn = -1;
 39         ch = getchar();
 40     }
 41     while (ch >= '0' && ch <= '9'){
 42         x = x * 10 + ch - '0';
 43         ch = getchar();
 44     }
 45     return sgn * x;
 46 }
 47  
 48 inline void add_edge(int x, int y, int v){
 49     e[++tot].next = first[x], first[x] = tot;
 50     e[tot].to = y, e[tot].v = v;
 51 }
 52  
 53 void add_Edges(const int x, const int y, const int v){
 54     add_edge(x, y, v);
 55     add_edge(y, x, v);
 56 }
 57  
 58 inline void add_to_heap(const int p){
 59     for (int x = first[p]; x; x = e[x].next)
 60         if (d[e[x].to] == -1){
 61             NODE.v = e[x].v + d[p], NODE.to = e[x].to;
 62             h.push(NODE);
 63         }
 64 }
 65  
 66 int Dijkstra(int S, int T){
 67     memset(d, -1, sizeof(d));
 68     while (!h.empty()) h.pop();
 69     d[S] = 0, add_to_heap(S);
 70     int p;
 71     while (d[T] == -1){
 72         while (d[h.top().to] != -1) h.pop();
 73         p = h.top().to;
 74         d[p] = h.top().v;
 75         h.pop();
 76         add_to_heap(p);
 77     }
 78     return d[T];
 79 }
 80  
 81 void build_graph(){
 82     int Cost, x, y;
 83     for (int i = 1; i <= n; ++i)
 84         for (int j = 1; j < m; ++j){
 85             Cost = read();
 86             x = i == 1 ? S : (2 * i - 3) * (m - 1) + j;
 87             y = i == n ? T : (2 * i - 2) * (m - 1) + j;
 88             add_Edges(x, y, Cost);
 89         }
 90     for (int i = 1; i < n; ++i)
 91         for (int j = 1; j <= m; ++j){
 92             Cost = read();
 93             x = j == 1 ? T : 2 * (i - 1) * (m - 1) + j - 1;
 94             y = j == m ? S : 2 * (i - 1) * (m - 1) + j - 1 + m;
 95             add_Edges(x, y, Cost);
 96         }
 97     for (int i = 1; i < n; ++i)
 98         for (int j = 1; j < m; ++j){
 99             Cost = read();
100             x = (2 * i - 2) * (m - 1) + j;
101             y = (2 * i - 1) * (m - 1) + j;
102             add_Edges(x, y, Cost);
103         }
104 }
105  
106 int main(){
107     n = read(), m = read();
108     if (n == 1 && m == 1){
109         printf("%d\n", 0);
110         return 0;
111     }
112      
113     S = 0, T = 2 * (m - 1) * (n - 1) + 1;
114     build_graph();         
115     printf("%d\n", Dijkstra(S, T));
116     return 0;
117 }
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posted on 2014-10-19 14:18  Xs酱~  阅读(280)  评论(0)  编辑  收藏  举报