hdu 4735Little Wish~ lyrical step~ 重复覆盖

题目链接

给出一棵树， 树上点的值为0或1， 可以交换树上两个点的权值， 给出一个距离m， 所有的0距离最近的1的距离不能超过m， 求最少的交换次数。

首先对于每一个点u，所有离u的距离不超过m的点v， 加一条边（u, v)。

然后dlx， 剪枝的函数是当前1的个数+还需要的1的个数不超过1的总数， 具体看代码。

#include<bits/stdc++.h>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, a, n) for(int i = a; i<n; i++)
#define ull unsigned long long
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 200;
const int maxNode = 10000;
int head[maxn], num;
struct node
{
    int to, nextt, c;
}e[maxn];
void addEdge(int u, int v, int c) {
    e[num].to = v;
    e[num].nextt = head[u];
    e[num].c = c;
    head[u] = num++;
}
struct DLX {
    int L[maxNode], R[maxNode], U[maxNode], D[maxNode], row[maxNode], col[maxNode];
    int S[maxn], H[maxn], deep, ans[maxn], sz, n, m, kind[maxn], boy_num;
    void remove(int c) {
        for(int i = D[c]; i!=c; i = D[i]) {
            L[R[i]] = L[i];
            R[L[i]] = R[i];
        }
    }
    void resume(int c) {
        for(int i = U[c]; i!=c; i = U[i]) {
            L[R[i]] = i;
            R[L[i]] = i;
        }
    }
    int h() {
        int cnt = 0;
        int vis[105];
        mem(vis);
        for(int i = R[0]; i!=0; i = R[i]) {
            if(!vis[i]) {
                cnt++;
                vis[i] = 1;
                for(int j = D[i]; j!=i; j = D[j]) {
                    for(int k = R[j]; k!=j; k = R[k]) {
                        vis[col[k]] = 1;
                    }
                }
            }
        }
        return cnt;
    }
    void dfs(int d, int cnt) {
        if(d+h()>boy_num||cnt>=deep)
            return ;
        if(R[0] == 0) {
            deep = min(deep, cnt);
            return ;
        }
        int c = R[0];
        for(int i = R[0]; i!=0; i = R[i])
            if(S[c]>S[i])
                c = i;
        for(int i = D[c]; i!=c; i = D[i]) {
            remove(i);
            for(int j = R[i]; j!=i; j = R[j])
                remove(j);
            dfs(d+1, cnt+(!kind[row[i]]));
            for(int j = L[i]; j!=i; j = L[j])
                resume(j);
            resume(i);
        }
        return ;
    }
    void add(int r, int c) {
        sz++;
        row[sz] = r;
        col[sz] = c;
        S[c]++;
        U[sz] = U[c];
        D[sz] = c;
        D[U[c]] = sz;
        U[c] = sz;
        if(~H[r]) {
            R[sz] = H[r];
            L[sz] = L[H[r]];
            L[R[sz]] = sz;
            R[L[sz]] = sz;
        } else {
            H[r] = L[sz] = R[sz] = sz;
        }
    }
    void init(){
        mem1(H);
        num = boy_num = 0;
        mem1(head);
        deep = inf;
        for(int i = 0; i<=n; i++) {
            R[i] = i+1;
            L[i] = i-1;
            U[i] = i;
            D[i] = i;
        }
        mem(S);
        R[n] = 0;
        L[0] = n;
        sz = n;
    }
    void dfs(int u, int root, int fa, int d) {
        add(root, u);
        for(int i = head[u]; ~i; i = e[i].nextt) {
            int v = e[i].to, c = e[i].c;
            if(v == fa)
                continue;
            if(c+d<=m)
                dfs(v, root, u, d+c);
        }
    }
    void solve() {
        init();
        int u, v, c;
        for(int i = 1; i<=n; i++) {
            scanf("%d", &kind[i]);
            boy_num+=kind[i];
        }
        for(int i = 0; i<n-1; i++) {
            scanf("%d%d%d", &u, &v, &c);
            addEdge(u, v, c);
            addEdge(v, u, c);
        }
        for(int i = 1; i<=n; i++) {
            dfs(i, i, -1, 0);
        }
        dfs(0, 0);
        if(deep == inf)
            deep = -1;
        cout<<deep<<endl;
    }
}dlx;
int main()
{
    int t;
    cin>>t;
    for(int i = 1; i<=t; i++) {
        printf("Case #%d: ", i);
        scanf("%d%d", &dlx.n, &dlx.m);
        dlx.solve();
    }
    return 0;
}