CSU 1541 there is No alternative (minimum spanning Tree + enumeration)

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Test instructions

There are n points, M edges, to make all n points together and to spend the least, ask which side is necessary to connect.

Analysis:

To minimize the cost of making sure is to make the smallest spanning tree, but the topic requires which sides are necessary to use. We can

To think like this, we first ask for the smallest spanning tree and then count the edges

Lift This n-1 the edges so they can't be used, and then continue to do the minimum spanning tree if the last request value and first

If not, then this side is sure to be connected.

The code is as follows:

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring>using namespace    std;const int MAXN = 50010;struct nod{int x,y,val;    BOOL operator < (const struct NOD &tmp) const{return this->val<tmp.val; }}edge[maxn];int par[maxn/100],num[maxn/100];int id[maxn/100];int n,m;void init () {for (int i=0;i<=n;i++) par[i]=i, Num[i]=1;}    int find_par (int x) {if (x!=par[x]) return Par[x]=find_par (Par[x]); return par[x];}    BOOL Union (int x,int y) {x=find_par (x);    Y=find_par (y);        if (x!=y) {par[y]=x;        Num[x]+=num[y];    return true; } return false;}    int ans, cnt,sum1,sum2;void solve () {sum1=sum2=0;        for (int i=0;i<cnt;i++) {int tmp=0;init (); for (int j=0;j<m;j++) {if (J!=id[i]) {if (Union (EDGE[J].X,EDGE[J].Y)) tmp+=ed            Ge[j].val;    }} if (Tmp!=ans) Sum1++,sum2+=edge[id[i]].val; } printf ("%d%d\n", sum1,sum2);} INT Main () {while (~scanf ("%d%d", &n,&m)) {init ();        for (int i=0;i<m;i++) {scanf ("%d%d%d", &edge[i].x,&edge[i].y,&edge[i].val);        } sort (edge,edge+m);        ans=0,cnt=0;        Memset (id,0,sizeof (id));                for (int i=0;i<m;i++) {if (Union (EDGE[I].X,EDGE[I].Y)) {ans+=edge[i].val;            Id[cnt++]=i;    }} solve (); } return 0;} /***4 41 2 31 3 52 3 32 4 34 41 2 31 3 12 3 32 4 33 31 2 12 3 11 3 1***/

CSU 1541 there is No alternative (minimum spanning Tree + enumeration)