uva 11248 最小割 

Dinic


  1 #include<iostream>
  2 #include<string>
  3 #include<algorithm>
  4 #include<cstdlib>
  5 #include<cstdio>
  6 #include<set>
  7 #include<map>
  8 #include<vector>
  9 #include<cstring>
 10 #include<stack>
 11 #include<cmath>
 12 #include<queue>
 13 using namespace std;
 14 #define INF 0x3f3f3f3f
 15 #define LL long long
 16 #define MAXN 110
 17 struct Edge{
 18     int from, to, cap, flow;
 19 };
 20 bool comp(const Edge& a, const Edge& b){
 21     return a.from < b.from || (a.from == b.from && a.to < b.to);
 22 }
 23 struct Dinic{
 24     int n, m, s, t;
 25     vector<Edge> edges;
 26     vector<int> G[MAXN];
 27     bool vis[MAXN];
 28     int d[MAXN];
 29     int cur[MAXN];
 30     void init(int n){
 31         this->n = n;
 32         for (int i = 0; i <= n; i++) 
 33             G[i].clear();
 34         edges.clear();
 35     }
 36     void AddEdge(int from, int to, int cap){
 37         edges.push_back(Edge{ from, to, cap, 0 });
 38         edges.push_back(Edge{ to, from, 0, 0 });
 39         m = edges.size();
 40         G[from].push_back(m - 2);
 41         G[to].push_back(m - 1);
 42     }
 43     bool BFS(){
 44         memset(vis, 0, sizeof(vis));
 45         queue<int> Q;
 46         Q.push(s);
 47         d[s] = 0;
 48         vis[s] = 1;
 49         while (!Q.empty()){
 50             int x = Q.front();
 51             Q.pop();
 52             for (int i = 0; i<G[x].size(); i++){
 53                 Edge& e = edges[G[x][i]];
 54                 if (!vis[e.to] && e.cap>e.flow){
 55                     vis[e.to] = 1;
 56                     d[e.to] = d[x] + 1;
 57                     Q.push(e.to);
 58                 }
 59             }
 60         }
 61         return vis[t];
 62     }
 63     int DFS(int x, int a){
 64         if (x == t || a == 0)return a;
 65         int flow = 0, f;
 66         for (int& i = cur[x]; i<G[x].size(); i++){
 67             Edge& e = edges[G[x][i]];
 68             if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow)))>0){
 69                 e.flow += f;
 70                 edges[G[x][i] ^ 1].flow -= f;
 71                 flow += f;
 72                 a -= f;
 73                 if (a == 0)break;
 74             }
 75         }
 76         return flow;
 77     }
 78     int Maxflow(int s, int t, int need){
 79         this->s = s; this->t = t;
 80         int flow = 0;
 81         while (BFS()){
 82             memset(cur, 0, sizeof(cur));
 83             flow += DFS(s, INF);
 84             if (flow > need)return flow;
 85         }
 86         return flow;
 87     }
 88     //最小割割边
 89     vector<int> Mincut(){
 90         BFS();
 91         vector<int> ans;
 92         for (int i = 0; i<edges.size(); i++){
 93             Edge& e = edges[i];
 94             if (vis[e.from] && !vis[e.to] && e.cap>0)ans.push_back(i);
 95         }
 96         return ans;
 97     }
 98     void Reduce(){
 99         for (int i = 0; i < edges.size(); i++) edges[i].cap -= edges[i].flow;
100     }
101     void ClearFlow(){
102         for (int i = 0; i < edges.size(); i++) edges[i].flow = 0;
103     }
104 };
105 Dinic ex;
106 int main(){
107     int N, E, C, cas = 0;
108     while (~scanf("%d%d%d", &N, &E, &C))
109     {
110         if (!N)break;
111         ex.init(N);
112         int a, b, c;
113         while (E--)
114         {
115             scanf("%d%d%d", &a, &b, &c);
116             ex.AddEdge(a, b, c);
117         }
118         int flow = ex.Maxflow(1, N, INF);
119         printf("Case %d: ", ++cas);
120         if (flow > C)printf("possible\n");
121         else{
122             vector<int> cut = ex.Mincut();
123             ex.Reduce();
124             vector<Edge>ans;
125             for (int i = 0; i < cut.size(); i++){
126                 Edge& e = ex.edges[cut[i]];
127                 int temp = e.cap;
128                 e.cap = C;
129                 ex.ClearFlow();
130                 if (flow + ex.Maxflow(1, N, C - flow) >= C)ans.push_back(e);
131                 e.cap = temp;
132             }
133             if (ans.empty())printf("not possible\n");
134             else{
135                 sort(ans.begin(), ans.end(), comp);
136                 printf("possible option:(%d,%d)", ans[0].from, ans[0].to);
137                 for (int i = 1; i < ans.size(); i++)
138                     printf(",(%d,%d)", ans[i].from, ans[i].to);
139                 printf("\n");
140             }
141         }
142     }
143     return 0;
144 }

 

posted on 2015-05-09 21:21  张济  阅读(136)  评论(0)    收藏  举报

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