UVA - 10480 Sabotage 最小割，输出割法

Posted 2021-01-09 mingsd

UVA - 10480 Sabotage

题意：现在有n个城市，m条路，现在要把整个图分成2部分，编号1，2的城市分成在一部分中，拆开每条路都需要花费，现在问达成目标的花费最少要隔开那几条路。

题解：建图直接按给你的图建一下，然后呢跑一下最大流，我们就知道了最小割是多少，答案就是最小割了  。

现在要求输出割法。我们从s开始往前跑，如果某条正向边有流量，我们就按着这条边继续往外走，知道无法再走，把所有经历过的点都染一下色。最后看所有的边，是不是有一头是染色了，另一头没有染色，如果是，这条边就是割边，输出就好了。

代码：

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
  4 #define LL long long
  5 #define ULL unsigned LL
  6 #define fi first
  7 #define se second
  8 #define pb emplace_back
  9 #define lson l,m,rt<<1
 10 #define rson m+1,r,rt<<1|1
 11 #define lch(x) tr[x].son[0]
 12 #define rch(x) tr[x].son[1]
 13 #define max3(a,b,c) max(a,max(b,c))
 14 #define min3(a,b,c) min(a,min(b,c))
 15 typedef pair<int,int> pll;
 16 const int inf = 0x3f3f3f3f;
 17 const LL INF = 0x3f3f3f3f3f3f3f3f;
 18 const LL mod =  (int)1e9+7;
 19 const int N = 200;
 20 const int M = N*N;
 21 int head[N], deep[N], cur[N];
 22 int w[M], to[M], nx[M];
 23 int tot;
 24 void add(int u, int v, int val){
 25     w[tot]  = val; to[tot] = v;
 26     nx[tot] = head[u]; head[u] = tot++;
 27 
 28     w[tot] = 0; to[tot] = u;
 29     nx[tot] = head[v]; head[v] = tot++;
 30 }
 31 int bfs(int s, int t){
 32     queue<int> q;
 33     memset(deep, 0, sizeof(deep));
 34     q.push(s);
 35     deep[s] = 1;
 36     while(!q.empty()){
 37         int u = q.front();
 38         q.pop();
 39         for(int i = head[u]; ~i; i = nx[i]){
 40             if(w[i] > 0 && deep[to[i]] == 0){
 41                 deep[to[i]] = deep[u] + 1;
 42                 q.push(to[i]);
 43             }
 44         }
 45     }
 46     return deep[t] > 0;
 47 }
 48 int Dfs(int u, int t, int flow){
 49     if(u == t) return flow;
 50     for(int &i = cur[u]; ~i; i = nx[i]){
 51         if(deep[u]+1 == deep[to[i]] && w[i] > 0){
 52             int di = Dfs(to[i], t, min(w[i], flow));
 53             if(di > 0){
 54                 w[i] -= di, w[i^1] += di;
 55                 return di;
 56             }
 57         }
 58     }
 59     return 0;
 60 }
 61 
 62 int n, m;
 63 int Dinic(int s, int t){
 64     int ans = 0, tmp;
 65     while(bfs(s, t)){
 66         for(int i = 1; i <= n; i++) cur[i] = head[i];
 67         while(tmp = Dfs(s, t, inf)) ans += tmp;
 68     }
 69     return ans;
 70 }
 71 int vis[N];
 72 int top;
 73 int ans[M][2];
 74 void init(){
 75     memset(head, -1, sizeof(head));
 76     memset(vis, 0, sizeof(vis));
 77     top = tot = 0;
 78 }
 79 void solve(int u){
 80     vis[u] = 1;
 81     for(int i = head[u]; ~i; i = nx[i]){
 82         if(vis[to[i]]) continue;
 83         if(w[i]){
 84             solve(to[i]);
 85         }
 86     }
 87 }
 88 int main(){
 89     while(~scanf("%d%d", &n, &m) && n + m){
 90         int u, v, w;
 91         init();
 92         for(int i = 1; i <= m; i++){
 93             scanf("%d%d%d", &u, &v, &w);
 94             add(u, v, w);
 95             add(v, u, w);
 96         }
 97         Dinic(1, 2);
 98         solve(1);
 99         for(int i = 1; i <= n; i++){
100             for(int j = head[i]; ~j; j = nx[j]){
101                 if(!(j&1)){
102                     v = to[j];
103                     if(vis[i] != vis[v] && i < v)
104                         printf("%d %d
", i, v);
105                 }
106             }
107         }
108         puts("");
109     }
110     return 0;
111 }

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