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  • HDU 3157 Crazy Circuits (有源汇上下界最小流)

    题意:一个电路板,上面有N个接线柱(标号1~N)   还有两个电源接线柱  +  - 然后是 给出M个部件正负极的接线柱和最小电流,求一个可以让所有部件正常工作的总电流。

    析:这是一个有源汇有上下界的最小流。

    有源汇有上下界最大流:

    1.构造附加网络

    2.对ss、tt求最大流(ss、tt满流则有解)

    3.若有解,对s、t求最大流

    有源汇有上下界最小流:

    1.构造附加网络(不添加[t,s]边)

    2.对ss、tt求最大流

    3.添加[t,s]边

    4.对ss、tt求最大流

    5.若ss、tt满流,则[t,s]的流量就是最小流

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    #include <sstream>
    #include <list>
    #include <assert.h>
    #include <bitset>
    #define debug() puts("++++");
    #define gcd(a, b) __gcd(a, b)
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define fi first
    #define se second
    #define pb push_back
    #define sqr(x) ((x)*(x))
    #define ms(a,b) memset(a, b, sizeof a)
    #define sz size()
    #define pu push_up
    #define pd push_down
    #define cl clear()
    #define all 1,n,1
    #define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const double inf = 1e20;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 80 + 10;
    const int maxm = 1e5 + 10;
    const int mod = 50007;
    const int dr[] = {-1, 0, 1, 0};
    const int dc[] = {0, -1, 0, 1};
    const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline bool is_in(int r, int c) {
      return r >= 0 && r < n && c >= 0 && c < m;
    }
    
    struct Edge{
      int from, to, cap, flow;
    };
    
    struct Dinic{
      int n, m, s, t;
      vector<Edge> edges;
      vector<int> G[maxn];
      int d[maxn];
      bool vis[maxn];
      int cur[maxn];
    
      void init(int n){
        this-> n = n;
        for(int i = 0; i < n; ++i)  G[i].cl;
        edges.cl;
      }
    
      void addEdge(int from, int to, int cap){
        edges.pb((Edge){from, to, cap, 0});
        edges.pb((Edge){to, from, 0, 0});
        m = edges.sz;
        G[from].pb(m-2);  G[to].pb(m-1);
      }
    
      bool bfs(){
        ms(vis, 0);  vis[s] = 1;
        d[s] = 1;
        queue<int> q;
        q.push(s);
    
        while(!q.empty()){
          int x = q.front();  q.pop();
          for(int i = 0; i < G[x].sz; ++i){
            Edge &e = edges[G[x][i]];
            if(!vis[e.to] && e.cap > e.flow){
              d[e.to] = d[x] + 1;
              vis[e.to] = 1;
              q.push(e.to);
            }
          }
        }
        return vis[t];
      }
    
      int dfs(int x, int a){
        if(x == t || a == 0)  return a;
        int flow = 0, f;
        for(int &i = cur[x]; i < G[x].sz; ++i){
          Edge &e = edges[G[x][i]];
          if(d[e.to] == d[x] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0){
            e.flow += f;
            edges[G[x][i]^1].flow -= f;
            a -= f;
            flow += f;
            if(a == 0)  break;
          }
        }
        return flow;
      }
    
      int maxflow(int s, int t){
        this->s = s;  this-> t = t;
        int flow = 0;
        while(bfs()){ ms(cur, 0);  flow += dfs(s, INF); }
        return flow;
      }
    
      void solve(int t){
        for(int i = 0; i < G[s].sz; ++i){
          Edge &e = edges[G[s][i]];
          if(e.cap > e.flow){
            puts("impossible");
            return ;
          }
        }
        printf("%d
    ", edges[*G[t].rbegin()].flow);
      }
    };
    
    Dinic dinic;
    int in[maxn];
    
    int main(){
      while(scanf("%d %d", &n, &m) == 2 && n+m){
        int s = n + 2, t = n + 3;
        dinic.init(n + 10);
        char a[15], b[15], c[15];
        ms(in, 0);
        for(int i = 0; i < m; ++i){
          scanf("%s %s %s", a, b, c);
          int u = a[0] == '+' ? 0 : atoi(a);
          int v = b[0] == '-' ? n+1 : atoi(b);
          int val = atoi(c);
          in[v] += val;
          in[u] -= val;
          dinic.addEdge(u, v, INF);
        }
        for(int i = 0; i <= n+1; ++i){
          if(in[i] > 0)  dinic.addEdge(s, i, in[i]);
          if(in[i] < 0)  dinic.addEdge(i, t, -in[i]);
        }
        dinic.maxflow(s, t);
        dinic.addEdge(n+1, 0, INF);
        dinic.maxflow(s, t);
        dinic.solve(n+1);
      }
      return 0;
    }
    

      

    Crazy Circuits

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/7629209.html
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