【codeforces 733F】Drivers Dissatisfaction

【题目链接】:http://codeforces.com/problemset/problem/733/F

【题意】

给你n个点m条边;
让你从中选出n-1条边;
形成一个生成树;
(即让n个点都联通);
然后,你有S的预算;
每次可以选择一条边i，然后花费ci的预算,把这条边的权值递减1;
(边一开始的权值为wi);
问你最后的最小生成树是多少;

【题解】

/*
    肯定是找某一条边一直减(ci最小的那一个,因为代价最小,又都是减少1);
    把m条边按照w升序排;
    做个最小生成树;
    把最小生成树里面c最小的那条边一直减就好,这个作为ans1;
    然后在这个最小生成树上写个dfs,
    搞出来每个点上面的第2^i个节点是谁,以及这个点到这个点上面的第2^i个点之间,最大的w是哪条边.
    然后枚举所有的非树边,这里的非树边,它的c的值一定要小于最小生成树里面的树边的最小的c值;
    这样它才有可能成为那条一直减小的边;然后最后比最小生成树里面的某条边边权来得小;
    假如非树边的两个点是u和v;
    则如果你要把u-v这条边加到MST里面的话,则必然要在从u到v的路径上选一条边删掉;
    删掉的边应是最大的那条边.(记录这条边是什么,删掉之后下次如果还要删的话,得还原)
    具体方式就是找到u和v的LCA;
    然后从u到lca的路径中找最大w的边;
    然后从v到lca的路径总找最大w的边;
    取两个边的w的较大者;
    然后把这条边删掉;
    然后把这条枚举的边加进去;一直减;然后更新答案;
*/

【Number Of WA】

0

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int M = 2e5+100;
const int INF = 21e8;

struct abc
{
    int w,c,u,v,id;
};

int n,m,f[M],b[M],fa[M][22],d[M][22],deep[M],C,num,bin[22],bb[M],pri[M];
LL tot,ans,S;
abc a[M];
vector <pii> g[M];

int ff(int x)
{
    if (f[x]==x)
        return x;
    else
        return f[x] = ff(f[x]);
}

int maxw(int x,int y)
{
    if (a[x].w>a[y].w)
        return x;
    else
        return y;
}

void dfs(int x)
{
    rep1(i,1,20)
        fa[x][i] = fa[fa[x][i-1]][i-1];
    rep1(i,1,20)
        d[x][i] = maxw(d[x][i-1],d[fa[x][i-1]][i-1]);
    for (pii temp:g[x])
    {
        int y = temp.fi,w = temp.se;
        if (y==fa[x][0]) continue;
        fa[y][0] = x,d[y][0] = w;
        deep[y] = deep[x]+1;
        dfs(y);
    }
}

int lcq(int x,int y)
{
    if (deep[x]<deep[y])
        swap(x,y);
    //deep[x]>=deep[y];
    int temp = deep[x]-deep[y];
    rep1(i,0,20)
        if (temp&bin[i])
            x = fa[x][i];
    //deep[x]==deep[y];
    rep2(i,20,0)
        if (fa[x][i]!=fa[y][i])
            x = fa[x][i],y = fa[y][i];
    return x==y?x:fa[x][0];
}

int query(int x,int y)
{
    //deep[x]>=deep[y]
    int temp = deep[x]-deep[y];
    int ret = 0;
    rep1(i,0,20)
        if (temp&bin[i])
        {
            ret = maxw(ret,d[x][i]);
            x = fa[x][i];
        }
    return ret;
}

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    ios::sync_with_stdio(false),cin.tie(0);//scanf,puts,printf not use
    //init??????
    bin[0] = 1;
    rep1(i,1,20) bin[i] = bin[i-1]<<1;
    cin >> n >> m;
    rep1(i,1,m) cin >> a[i].w;
    rep1(i,1,m) cin >> a[i].c;
    rep1(i,1,m)
    {
        cin >> a[i].u >> a[i].v;
        a[i].id = i;
    }
    cin >>S;
    sort(a+1,a+1+m,[&](abc a,abc b) {return a.w<b.w;});
    C = INF;
    rep1(i,1,n) f[i] = i;
    rep1(i,1,m)
    {
        int x = a[i].u,y = a[i].v;
        int r1 = ff(x),r2 = ff(y);
        if (r1!=r2)
        {
            f[r1] = r2;
            g[x].pb(mp(y,i));
            g[y].pb(mp(x,i));
            b[i] = 1;
            if (a[i].c<C)
            {
                C = a[i].c;
                num = i;
            }
            tot+=a[i].w;
        }
    }
    ans = tot-S/C;
    dfs(1);
    int dl = 0,prei = 0;
    rep1(i,1,m)
        if (a[i].c<C)
        {
            int x = a[i].u,y = a[i].v;
            int z = lcq(x,y);
            int tmp = maxw(query(x,z),query(y,z));
            LL temp = tot-a[tmp].w+a[i].w;
            temp-=S/a[i].c;
            if (temp<ans)
            {
                b[dl] = 1;b[dl=tmp] = 0;
                b[prei] = 0;prei = i;
                num = i;b[i] = 1;
                ans = temp;
            }
        }
    cout << ans << endl;
    rep1(i,1,m)
        if (b[i])
        {
            bb[a[i].id] = 1;
            if (num==i)
                pri[a[i].id] = a[i].w-S/a[i].c;
            else
                pri[a[i].id] = a[i].w;
        }
    rep1(i,1,m)
        if (bb[i])
            cout << i <<' '<<pri[i]<<endl;
    return 0;
}