码迷,mamicode.com
首页 > 其他好文 > 详细

[BeiJing2006]狼抓兔子

时间:2018-08-15 10:36:59      阅读:161      评论:0      收藏:0      [点我收藏+]

标签:rom   one   main   cap   ems   www   jin   efi   puts   

题面

 

一眼看就是最小割板子题,建图也很直观,注意每一条边建双向边其实不用建4条边,只要反向边的容量和正边相同就行。然后直接跑最大流板子就行。不过此题拿vector存图会MLE……而拿链前存图就能卡过去……场面一度十分尴尬。

这里发一个vector80分代码,各位改成链前就能AC了……

技术分享图片
  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cmath>
  4 #include<algorithm>
  5 #include<cstring>
  6 #include<cstdlib>
  7 #include<queue>
  8 #include<stack>
  9 #include<vector>
 10 #include<cctype>
 11 using namespace std;
 12 #define enter puts("")
 13 #define space putchar(‘ ‘)
 14 #define Mem(a) memset(a, 0, sizeof(a))
 15 #define rg register
 16 typedef long long ll;
 17 typedef double db;
 18 const int INF = 0x3f3f3f3f;
 19 const db eps = 1e-8;
 20 const int maxn = 1e6 + 5;
 21 inline ll read()
 22 {
 23     ll ans = 0;
 24     char ch = getchar(), last =  ;
 25     while(!isdigit(ch)) {last = ch; ch = getchar();}
 26     while(isdigit(ch))
 27     {
 28         ans = ans * 10 + ch - 0; ch = getchar();
 29     }
 30     if(last == -) ans = -ans;
 31     return ans;
 32 }
 33 inline void write(ll x)
 34 {
 35     if(x < 0) putchar(-), x = -x;
 36     if(x >= 10) write(x / 10);
 37     putchar(x % 10 + 0);
 38 }
 39 
 40 int n, m;
 41 
 42 struct Edge
 43 {
 44     int from, to, cap, flow;
 45 };
 46 vector<Edge> edges;
 47 vector<int> G[maxn];
 48 void addedge(int from, int to, int w)
 49 {
 50     edges.push_back((Edge){from, to, w, 0});
 51     edges.push_back((Edge){to, from, w, 0});
 52     int sz = edges.size();
 53     G[from].push_back(sz - 2);
 54     G[to].push_back(sz - 1);
 55 }
 56 
 57 int dis[maxn];
 58 bool vis[maxn];
 59 bool bfs()
 60 {
 61     for(rg int i = 1; i <= n * m; ++i) vis[i] = 0;
 62     dis[1] = 0; vis[1] = 1;
 63     queue<int> q; q.push(1);
 64     while(!q.empty())
 65     {
 66         int now = q.front(); q.pop();
 67         for(rg int i = 0; i < (int)G[now].size(); ++i)
 68         {
 69             Edge& e = edges[G[now][i]];
 70             if(!vis[e.to] && e.cap > e.flow)
 71             {
 72                 vis[e.to] = 1;
 73                 dis[e.to] = dis[now] + 1;
 74                 q.push(e.to);
 75             }
 76         }
 77     }
 78     return vis[n * m];
 79 }
 80 int cur[maxn];
 81 int dfs(const int& now, int a)
 82 {
 83     if(now == n * m || !a) return a;
 84     int flow = 0, f;
 85     for(int& i = cur[now]; i < (int)G[now].size(); ++i)
 86     {
 87         Edge& e = edges[G[now][i]];
 88         if(dis[e.to] == dis[now] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0)
 89         {
 90             e.flow += f;
 91             edges[G[now][i] ^ 1].flow -= f;
 92             flow += f; a -= f;
 93             if(!a) break;
 94         }
 95     }
 96     return flow;
 97 }
 98 
 99 int maxflow()
100 {
101     int flow = 0;
102     while(bfs())
103     {
104         for(rg int i = 1; i <= n * m; ++i) cur[i] = 0;
105         flow += dfs(1, INF);
106     }
107     return flow;
108 }
109 
110 int main()
111 {
112     n = read(); m = read();
113     for(rg int i = 1; i <= n; ++i)
114         for(rg int j = 1; j < m; ++j)
115         {
116             int x = read(), from = (i - 1) * m + j, to = (i - 1) * m + j + 1;
117             addedge(from, to, x);
118         }
119     for(rg int i = 1; i < n; ++i)
120         for(rg int j = 1; j <= m; ++j)
121         {
122             int x = read(), from = (i - 1) * m + j, to = i * m + j;
123             addedge(from, to, x);
124         }
125     for(rg int i = 1; i < n; ++i)
126         for(rg int j = 1; j < m; ++j)
127         {
128             int x = read(), from = (i - 1) * m + j, to = i * m + j + 1;
129             addedge(from, to, x);
130         }
131     write(maxflow()); enter;
132     return 0;
133 }
View Code

 

然后此题据说也可以用对偶图的知识解决。

[BeiJing2006]狼抓兔子

标签:rom   one   main   cap   ems   www   jin   efi   puts   

原文地址:https://www.cnblogs.com/mrclr/p/9479589.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!