Barricade

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1117 Accepted Submission(s): 340

Problem Description

The empire is under attack again. The general of empire is planning to defend his castle. The land can be seen as

Input

The first line of input contains an integer

Output

For each test cases, output the minimum wood cost.

Sample Input

1 4 4 1 2 1 2 4 2 3 1 3 4 3 4

Sample Output

最短路+网络流。

先一遍bfs找到最短路，再一次bfs找到最短路上的点，通过dis[i]+1 = dis[u]来找，然后跑一遍Dinic。

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #include <vector>
  6 #include <queue>
  7 using namespace std;
  8 const int maxn = 1005;
  9 const int inf = 0x3f3f3f3f;
 10 int n,m;
 11 int g[maxn][maxn];
 12 int vis[maxn];
 13 int dis[maxn];
 14 struct edge
 15 {
 16     int to;
 17     int cap;
 18     int rev;
 19 };
 20 vector<edge> gg[maxn];
 21 int level[maxn];
 22 int it[maxn];
 23 void add(int from,int to,int cap)
 24 {
 25     edge cur;
 26     cur.to = to;
 27     cur.cap = cap;
 28     cur.rev = gg[to].size();
 29     gg[from].push_back(cur);
 30     cur.to = from;
 31     cur.cap = 0;
 32     cur.rev = gg[from].size()-1;
 33     gg[to].push_back(cur);
 34 }
 35 
 36 void bfs(int s)
 37 {
 38     memset(level,-1,sizeof(level));
 39     queue<int> q;
 40     level[s] = 0;
 41     q.push(s);
 42     while(!q.empty())
 43     {
 44         int v = q.front(); q.pop();
 45         for(int i=0;i<gg[v].size();i++)
 46         {
 47             edge &e = gg[v][i];
 48             if(e.cap>0&&level[e.to]<0)
 49             {
 50                 level[e.to] = level[v]+1;
 51                 q.push(e.to);
 52             }
 53         }
 54     }
 55 }
 56 int dfs(int v,int t,int f)
 57 {
 58     if(v==t) return f;
 59     for(int &i=it[v];i<gg[v].size();i++)
 60     {
 61         edge &e = gg[v][i];
 62         if(e.cap>0&&level[v]<level[e.to])
 63         {
 64             int d = dfs(e.to,t,min(f,e.cap));
 65             if(d>0)
 66             {
 67                 e.cap -= d;
 68                 gg[e.to][e.rev].cap += d;
 69                 return d;
 70             }
 71         }
 72     }
 73     return 0;
 74 }
 75 int max_flow(int s,int t)
 76 {
 77     int flow = 0;
 78     for(;;)
 79     {
 80         bfs(s);
 81         if(level[t]<0) return flow;
 82         memset(it,0,sizeof(it));
 83         int f;
 84         while((f=dfs(s,t,inf))>0) flow += f;
 85     }
 86 }
 87 bool bfs1()
 88 {
 89     queue<int> q;
 90     memset(vis,0,sizeof(vis));
 91     memset(dis,inf,sizeof(dis));
 92     vis[1] = 1;
 93     dis[1] = 0;
 94     q.push(1);
 95     while(!q.empty())
 96     {
 97         int cur = q.front();q.pop();
 98         if(cur==n) return true;
 99         for(int i=1;i<=n;i++)
100         {
101             if(cur==i) continue;
102             if(!vis[i]&&g[cur][i]!=-1)
103             {
104                 vis[i] = 1;
105                 dis[i] = dis[cur]+1;
106                 q.push(i);
107             }
108         }
109     }
110     return false;
111 }
112 void bfs2()
113 {
114     queue<int> q;
115     memset(vis,0,sizeof(vis));
116     vis[n] = 1;
117     q.push(n);
118     while(!q.empty())
119     {
120         int cur = q.front();q.pop();
121         for(int i=1;i<=n;i++)
122         {
123             if(cur==i) continue;
124             if(g[cur][i]==-1) continue;
125             if(dis[i]+1==dis[cur])
126             {
127                 add(i,cur,g[i][cur]);
128                 if(!vis[i])
129                 {
130                     vis[i] = 1;
131                     q.push(i);
132                 }
133             }
134         }
135     }
136 }
137 int main()
138 {
139     int T;cin>>T;
140     while(T--)
141     {
142         scanf("%d %d",&n,&m);
143         int u,v,w;
144         memset(g,-1,sizeof(g));
145         for(int i=0;i<maxn;i++) gg[i].clear();
146         for(int i=1;i<=m;i++)
147         {
148             scanf("%d %d %d",&u,&v,&w);
149             g[u][v] = w;
150             g[v][u] = w;
151         }
152         int ans = 0;
153         bfs1();
154         bfs2();
155         ans = max_flow(1,n);
156         printf("%d\n",ans);
157     }
158     return 0;
159 }

HDU 5889 (最短路+网络流)

Barricade

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