一.题目链接:http://poj.org/problem?id=1861
二.题目大意:给若干边,有起点,终点,和权重。求能走完所有边的一个子图,要求单个边是最小的。
三.思路:最小生成树的求法,求完之后最后一条边肯定是最大的边,并且也满足所有的点都走一遍。(反证法:如果最小生成树最后的一条边不是能达到的最小边,假设有边比它小,那么出队列的就不会是它)。
四.代码:
#include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <cstring> #include <queue> using namespace std; const int MAX_SIZE = 1002, INF = 1<<30, MOD = 1000000007; struct Edge { int u, v, weight; }; Edge edges[MAX_SIZE*MAX_SIZE]; int nodeNum, edgeNum, pre[MAX_SIZE], pos1[MAX_SIZE], pos2[MAX_SIZE]; //pos1,pos2保存以连接的2个点。 void FUset() { memset(pre, -1, sizeof(pre)); } int Find(int x) { int root = x, save; while(pre[root] >= 0){ root = pre[root]; } while(x != root){ save = pre[x]; pre[x] = root; x = save; } return root; } void Union(int x, int y) { int xRoot = Find(x), yRoot = Find(y); //此时temp必为负数,负多少表示这棵树有几个节点 int temp = pre[xRoot] + pre[yRoot]; //把节点少的树合并在节点多的树下面。 if(pre[xRoot] > pre[yRoot]){ pre[xRoot] = yRoot; pre[yRoot] = temp; } else{ pre[yRoot] = xRoot; pre[xRoot] = temp; } } bool cmp(Edge a, Edge b) { return a.weight < b.weight; } int Kruskal(int &buildEdgeNum) { int i, u, v, maxWeight; sort(edges, edges + edgeNum, cmp); FUset(); buildEdgeNum = 0; for(i = 0; i < edgeNum; i++){ u = Find(edges[i].u); v = Find(edges[i].v); if(u != v){ maxWeight = edges[i].weight; Union(u, v); pos1[buildEdgeNum] = edges[i].u; pos2[buildEdgeNum] = edges[i].v; buildEdgeNum++; } if(buildEdgeNum >= nodeNum - 1) break; } return maxWeight; } int main() { //freopen("in.txt", "r", stdin); int i, j, test = 1, buildEdgeNum; while(cin>>nodeNum>>edgeNum){ for(i = 0; i < edgeNum; i++) cin>>edges[i].u>>edges[i].v>>edges[i].weight; cout<<Kruskal(buildEdgeNum)<<endl; cout<<buildEdgeNum<<endl; for(i = 0; i < buildEdgeNum; i++) cout<<pos1[i]<<" "<<pos2[i]<<endl; } }