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-;
 const int mod = 1e9+;
 const int inf = ;
 const int dir[][] = { {-, }, {, }, {, -}, {, } };
 const int maxn = ;
 const int maxNode = ;
 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 = ;
         int vis[];
         mem(vis);
         for(int i = R[]; i!=; i = R[i]) {
             if(!vis[i]) {
                 cnt++;
                 vis[i] = ;
                 for(int j = D[i]; j!=i; j = D[j]) {
                     for(int k = R[j]; k!=j; k = R[k]) {
                         vis[col[k]] = ;
                     }
                 }
             }
         }
         return cnt;
     }
     void dfs(int d, int cnt) {
         if(d+h()>boy_num||cnt>=deep)
             return ;
         if(R[] == ) {
             deep = min(deep, cnt);
             return ;
         }
         int c = R[];
         for(int i = R[]; i!=; 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+, 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 = ;
         mem1(head);
         deep = inf;
         for(int i = ; i<=n; i++) {
             R[i] = i+;
             L[i] = i-;
             U[i] = i;
             D[i] = i;
         }
         mem(S);
         R[n] = ;
         L[] = 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 = ; i<=n; i++) {
             scanf("%d", &kind[i]);
             boy_num+=kind[i];
         }
         for(int i = ; i<n-; i++) {
             scanf("%d%d%d", &u, &v, &c);
             addEdge(u, v, c);
             addEdge(v, u, c);
         }
         for(int i = ; i<=n; i++) {
             dfs(i, i, -, );
         }
         dfs(, );
         if(deep == inf)
             deep = -;
         cout<<deep<<endl;
     }
 }dlx;
 int main()
 {
     int t;
     cin>>t;
     for(int i = ; i<=t; i++) {
         printf("Case #%d: ", i);
         scanf("%d%d", &dlx.n, &dlx.m);
         dlx.solve();
     }
     return ;
 }