POJ1861 Network

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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;    }}

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