pat-top 1013. Image Segmentation (35)

https://www.patest.cn/contests/pat-t-practise/1013

这道题自己的想法是 算出一个个mst，然后看每个MST的最长边是否可以分解。 但是没想出好的dfs模型

后来看了http://blog.csdn.net/jtjy568805874/article/details/53435235，逆向思维，从搭建compoent的角度出发。去填边，正是堆+并查集求MST的思路，很精彩。

code from http://blog.csdn.net/jtjy568805874/article/details/53435235

#include <cstdio>#include <algorithm>#include <vector>using namespace std;#define rep(i,j,k) for(int i=j;i<=k;i++)#define inthr(x,y,z) scanf("%d%d%lf",&x,&y,&z)const int maxn = 8e3 + 10;int n, m;int fa[maxn], c[maxn]; //fa:father;  nodes count of componentdouble C, mw[maxn]; //mw:max edge weight;vector<int> res[maxn];// restore the componentstypedef struct edge {int x, y; double w;bool friend operator < (struct edge a, struct edge b) {return a.w < b.w;}}Edge;Edge edges[maxn];int find(int x) {if (x == fa[x]) return x;fa[x] = find(fa[x]);c[fa[x]] += c[x]; c[x] = 0; //merge the nodes countreturn fa[x];}int cmp(vector<int> a,vector<int> b) {if (!a.size() && b.size()) return true;if (!b.size()) return false;return a[0] < b[0];}int main(){inthr(n, m, C);rep(i, 0, n - 1) fa[i] = i, mw[i] = 0, c[i] = 1;rep(i, 0, m - 1) inthr(edges[i].x, edges[i].y, edges[i].w);sort(edges, edges + m);rep(i, 0, m - 1) {int fx = find(edges[i].x), fy = find(edges[i].y);if (fx == fy) continue;if (fx > fy) swap(fx, fy);if (mw[fx] + C / c[fx] < edges[i].w) continue;if (mw[fy] + C / c[fy] < edges[i].w) continue;fa[fy] = fx; find(fy); mw[fx] = edges[i].w;}rep(i, 0, n - 1) res[find(i)].push_back(i);rep(i, 0, n - 1) sort(res[i].begin(), res[i].end());sort(res, res + n, cmp);rep(i, 0, n - 1){if (res[i].size()){rep(j, 0, res[i].size() - 1){printf("%s%d",j==0?"":" ",res[i][j]);}printf("\n");}}return 0;}