hdu 4735Little Wish~ lyrical step~ 重复覆盖

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 ;

 }

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hdu 4735Little Wish~ lyrical step~ 重复覆盖

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