HDU 4700 Flow（瓶颈生成树）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4700

题意：给出一张N个点的图的最小割矩阵，问能否构造出一个满足该矩阵的图

思路：这题和之前长春区域赛的一道题有相似之处，考虑最终构造出一棵树，先将边降序排列，这样可以保证每次新加进去的边都是该生成树的瓶颈路，而瓶颈路恰限制住了两点间的流量，不过原图如果存在g[i][j] < min(g[i][k], g[k][j])的情况则该图无法构出

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <climits>
#include <functional>
#include <deque>
#include <ctime>
#include <string>

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#pragma comment(linker, "/STACK:102400000,102400000")

using namespace std;

const int maxn = 105;

struct edge
{
int u, v, w;
bool operator < (const edge &rhs) const
{
return w > rhs.w;
}
};

int fa[maxn];
int a[maxn][maxn];
int ans[maxn][maxn];

int Find(int x)
{
if (x == fa[x]) return x;
return fa[x] = Find(fa[x]);
}

bool unite(int x, int y)
{
int fx = Find(x);
int fy = Find(y);
if (fx == fy)
return false;
fa[fx] = fy;
return true;
}

vector <edge> g;

int main()
{
int n;
while (~scanf("%d", &n))
{
g.clear();
memset(ans, 0, sizeof(ans));
for (int i = 1; i <= n; i++)
{
fa[i] = i;
ans[i][i] = -1;
for (int j = 1; j <= n; j++)
{
edge e;
e.u = i, e.v = j;
scanf("%d", &e.w);
g.push_back(e);
a[i][j] = e.w;
}
}

int fl = 0;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
for (int k = 1; k <= n; k++)
{
if (i == j || j == k || i == k) continue;
if (a[i][j] < min(a[i][k], a[k][j]))
fl = 1;
}

if (fl)
{
puts("NO");
continue;
}
sort(g.begin(), g.end());
// for(int i = 0; i < g.size(); i++)
// printf("%d %d %d\n", g[i].u, g[i].v, g[i].w);

for (int i = 0; i < g.size(); i++)
{
edge e = g[i];
int u = e.u, v = e.v;
if (unite(u, v))
ans[u][v] = ans[v][u] = e.w;
}

puts("YES");
for (int i = 1; i <= n; i++)
{
printf("%d", ans[i][1]);
for (int j = 2; j <= n; j++)
{
printf(" %d", ans[i][j]);
}
printf("\n");
}
}
return 0;
}