HDU 4700 flow Gomory–Hu tree,通過已知最大流構造圖形

題目連結  http://acm.hdu.edu.cn/showproblem.php?pid=4700

Gomory–Hu tree http://en.wikipedia.org/wiki/Gomory%E2%80%93Hu_tree

Gomory–Hu tree的定義

Let G = ((VG, EG), c) be an undirected graph withc(u,v) being the capacity of the edge (u,v) respectively.

Denote the minimum capacity of an s- t cut by λst for each s,
t ∈ VG.

Let T = ( VT, ET) be a tree with VT =
VG, denote the set of edges in an s- t path by
Pst for each s, t ∈ VT.

Then T is said to be a Gomory–Hu tree of G if

λst = mine∈Pst c( Se, Te) for all
s, t ∈ VG,

where

1. Se and Te are the two connected components ofT∖{e} in the sense that (Se,
   Te) form as-t cut in G, and
2. c(Se, Te) is the capacity of the cut inG.

通俗來講就是對任意的  兩個節點s,t，Gomory-Hu tree 儲存的是他們各自相鄰節點的最大流。

要求兩個點的 最大流的話，就是求 從 s點沿著樹的邊走到t，其中的最小的邊的權值。

此題的話，標準題解裡的解釋是：  如果存在這樣要求的一副圖的話，必然存在它所對應的 Gomory–Hu tree ，

而把Gomory–Hu tree的非相鄰邊設定為0，也就是可以成立的一個答案了。

標準題解:

遞迴構造

令 f∗ = mina̸=b Fa,b
令 A = {v0 } ∪ {v : Fv,v0 > f∗ }, B = {v : Fv,v0 = f∗ }。

若 B = ∅,失敗
若 ∃(a ∈ A, b ∈ B)Fa,b > f∗ ,失敗
遞迴構造 A, B
取 a∗ ∈ A, b∗ ∈ B,加邊 (a∗ , b∗ ),容量是 f∗

代碼:

#include <cassert>#include <cstdio>#include <cstring>#include <climits>#include <vector>#include <algorithm>#define SIZE(v) ((int)((v).size()))#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)const int N = 100;int n, cut[N][N], graph[N][N];bool check(std::vector <int> vs) {    if (SIZE(vs) <= 1) {        return true;    }    int bridge = INT_MAX;    foreach (u, vs) {        foreach (v, vs) {            bridge = std::min(bridge, cut[*u][*v]);        }    }    int u0 = vs.front();    std::vector <int> v0, v1;    foreach (iter, vs) {        int v = *iter;        if (cut[u0][v] > bridge) {            v0.push_back(v);        } else {            v1.push_back(v);        }    }    if (v0.empty() || v1.empty()) {        return false;    }    graph[v0.front()][v1.front()] = graph[v1.front()][v0.front()] = bridge;    foreach (u, v0) {    //兩個集合 v0跟v1中的最大流是 bridge,如果給出的條件中 flow(a,b) > bridge，則與條件矛盾，構造不出該圖形        foreach (v, v1) {            if (cut[*u][*v] != bridge) {                return false;            }        }    }    return check(v0) && check(v1);}int main() {//freopen("1005.in","r",stdin);//freopen("10050.out","w",stdout);    freopen("temp.in","r",stdin);    while (scanf("%d", &n) == 1) {    //printf("%d\n",n);        assert(1 <= n && n <= N);        for (int i = 0; i < n; ++ i) {            for (int j = 0; j < n; ++ j) {                assert(scanf("%d", &cut[i][j]) == 1);             //   if(j!=n-1)printf("%d ",cut[i][j]);             //   else printf("%d",cut[i][j]);                if (i == j) {                    assert(cut[i][j] == -1);                    cut[i][j] = INT_MAX;                } else {                    assert(0 <= cut[i][j] && cut[i][j] <= 1000000000);                }            }        //    puts("");        }        for (int i = 0; i < n; ++ i) {            for (int j = 0; j < n; ++ j) {                assert(cut[i][j] == cut[j][i]);            }        }        std::vector <int> vertices;        for (int i = 0; i < n; ++ i) {            vertices.push_back(i);        }        memset(graph, 0, sizeof(graph));        for (int i = 0; i < n; ++ i) {            graph[i][i] = -1;        }        if (check(vertices)) {            puts("YES");            for (int i = 0; i < n; ++ i) {                for (int j = 0; j < n; ++ j) {                    //printf("%d%c", i == j ? -1 : 0, j == n - 1 ? '\n' : ' ');                    printf("%d%c", graph[i][j], j == n - 1 ? '\n' : ' ');                }            }        } else {            puts("NO");        }    }    //puts(check(vertices) ? "YES" : "NO");    return 0;}