hdu 4603 Color the Tree
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这道题细节真的非常多
首先能够想到a和b的最优策略一定是沿着a和b在树上的链走,走到某个点停止,然后再依次占据和这个点邻接的边
所以,解决这道题的过程例如以下:
预处理阶段:
step 1:取随意一个点为根节点。找出父子关系而且对这个树进行dp。求出从某个节点出发往下所包括的全部边的权值总和 复杂度O(n)
step 2:从tree dp 的结果中计算对于某个节点。从某条边出发所包括的边的综合。而且对其从大到小进行排序 复杂度O(n*logn)
step 3:dfs求出这颗树的欧拉回路,以及每一个点的dfn,而且按欧拉回路的顺序计算每一个节点的深度 复杂度O(2*n)
step 4:利用sparse table算法初始化step 3中的深度序列 复杂度 O(n*logn)
step 5:计算出从某个节点往上走2的n次方步所到达的节点 复杂度O(n*logn)
查询阶段:
关键是找到两点的 LCA 以及相遇点,而且找到一条或两条所经过且和相遇点邻接的边
分几种情况讨论
1. 两个点在一起
2.两个点之间的距离为1
3.dep[a] == dep[b]
4.dep[a] > dep[b] + 1
5.dep[a] < dep[b]
6.dep[a] == dep[b]+1
ps:少考虑第六种情况wa了一个下午
#pragma comment(linker, "/STACK:102400000,102400000") #include<iostream> #include<vector> #include<algorithm> #include<cstdio> #include<queue> #include<stack> #include<string> #include<map> #include<set> #include<cmath> #include<cassert> #include<cstring> #include<iomanip> #include<ctime> using namespace std; #ifdef _WIN32 typedef __int64 i64; #define out64 "%I64d\n" #define in64 "%I64d" #else typedef long long i64; #define out64 "%lld\n" #define in64 "%lld" #endif /************ for topcoder by zz1215 *******************/ #define foreach(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++) #define FOR(i,a,b) for( int i = (a) ; i <= (b) ; i ++) #define FF(i,a) for( int i = 0 ; i < (a) ; i ++) #define FFD(i,a,b) for( int i = (a) ; i >= (b) ; i --) #define S64(a) scanf(in64,&a) #define SS(a) scanf("%d",&a) #define LL(a) ((a)<<1) #define RR(a) (((a)<<1)+1) #define pb push_back #define pf push_front #define X first #define Y second #define CL(Q) while(!Q.empty())Q.pop() #define MM(name,what) memset(name,what,sizeof(name)) #define MC(a,b) memcpy(a,b,sizeof(b)) #define MAX(a,b) ((a)>(b)?(a):(b)) #define MIN(a,b) ((a)<(b)?(a):(b)) #define read freopen("in.txt","r",stdin) #define write freopen("out.txt","w",stdout) const int inf = 0x3f3f3f3f; const long long inf64 = 0x3f3f3f3f3f3f3f3fLL; const double oo = 10e9; const double eps = 10e-9; const double pi = acos(-1.0); const int maxn = 100111; const int maxlevel = 21; struct Node { int now; int to; int c; int tot; int ss; const bool operator < (const Node& cmp) const { return tot > cmp.tot; } }; int all; int n, m; vector<Node>g[maxn]; int t[maxn]; int dep[maxn]; int df; int dfn[maxn]; int dfv[maxn * 2]; int st[maxn * 2][maxlevel]; int up[maxn][maxlevel]; int dp[maxn]; //down sum int xtof[maxn]; int ftox[maxn]; int vis[maxn]; int lg2[maxn*2]; void dfs(int now) { vis[now] = true; int to; for (int i = 0; i < (int)g[now].size(); i++) { to = g[now][i].to; if (!vis[to]) { t[to] = now; dfs(to); } } } int treedp(int now) { int to,id; dp[now] = 0; for (int i = 0; i < (int)g[now].size(); i++) { to = g[now][i].to; if (to != t[now]) { int temp = treedp(to) + g[now][i].c; g[now][i].tot = temp; dp[now] += temp; } else { id = i; } } if (t[now] != -1) { g[now][id].tot = all - dp[now]; } return dp[now]; } void euler_circuit(int now ,int step) { dep[now] = step; dfn[now] = df; dfv[df++] = now; int to; for (int i = 0; i < (int)g[now].size(); i++) { to = g[now][i].to; if (to != t[now]) { euler_circuit(to,step+1); dfv[df++] = now; } } } void get_up_node() { for (int i = 1; i <= n; i++) { up[i][0] = t[i]; } int to; for (int step = 1; step < maxlevel; step++) { for (int now = 1; now <= n; now++) { to = up[now][step - 1]; if (to == -1) { up[now][step] = -1; } else { up[now][step] = up[to][step - 1]; } } } } void sparse_table() { for (int i = 1; i < df; i++){ st[i][0] = dep[dfv[i]]; } int to; for (int step = 1; step <= lg2[n] + 1; step++){ for (int now = 1; now < df; now++) { to = now + (1 << (step - 1)); if (to < df){ st[now][step] = min(st[now][step - 1], st[to][step - 1]); } else{ st[now][step] = st[now][step - 1]; } } } } void relation() { int to; for (int now = 1; now <= n; now++){ for (int i = 0; i < (int)g[now].size(); i++){ to = g[now][i].to; if (to == t[now]){ xtof[now] = i; } else{ ftox[to] = i; } } } } int rmq(int l,int r) { return min(st[l][lg2[r - l]], st[r - (1 << lg2[r - l])][lg2[r - l]] ); } int calculate(int x,bool first,int id1,int id2=-1) { if (id2 != -1){ if (id1 > id2){ swap(id1, id2); } } int sum = g[x][0].ss; sum -= g[x][id1].tot; if (id2 != -1){ sum -= g[x][id2].tot; } int size = (int)g[x].size() - 1; if (size >= 1){ sum += g[x][1].ss; } int ans = g[x][0].ss; if (id1 % 2 ){ if (id1 + 1 <= size){ ans -= g[x][id1 + 1].ss; if (id1 + 2 <= size){ ans += g[x][id1 + 2].ss; } } if (id2 != -1){ if (id2 % 2){ ans -= g[x][id2].ss; if (id2 + 1 <= size){ ans += g[x][id2 + 1].ss; } } else{ if (id2 + 1 <= size){ ans -= g[x][id2 + 1].ss; if (id2 + 2 <= size){ ans += g[x][id2 + 2].ss; } } } } } else{ ans -= g[x][id1].ss; if (id1 + 1 <= size){ ans += g[x][id1 + 1].ss; } if (id2 != -1){ if (id2 % 2){ ans -= g[x][id2].ss; if (id2 + 1 <= size){ ans += g[x][id2 + 1].ss; } } else{ if (id2 + 1 <= size){ ans -= g[x][id2 + 1].ss; if (id2 + 2 <= size){ ans += g[x][id2 + 2].ss; } } } } } if (first) return ans; else return sum - ans; } int go_up(int now, int x) { int step = 0; while (x) { if (x & 1) { now = up[now][step]; } step++; x >>= 1; } return now; } int find(int a,int b) { int l = dfn[a]; int r = dfn[b]; if (l == r){ return g[a][0].ss; } if (l > r){ swap(l, r); } int lca = rmq(l, r + 1); //dep if (dep[a] - lca + dep[b] - lca == 1){ if (dep[a] == lca){ return g[b][xtof[b]].tot + calculate(b, false, xtof[b]); } else if (dep[b] == lca){ return g[b][ftox[a]].tot + calculate(b, false, ftox[a]); } } else if (dep[a] > dep[b]+1){ int temp = dep[a] - dep[b]; int mid = lca + temp / 2; int child = go_up(a, dep[a] - mid - 1); if (temp % 2){ return g[t[child]][ftox[child]].tot + calculate(t[child], false, ftox[child], xtof[t[child]]); } else{ return g[t[child]][ftox[child]].tot + calculate(t[child], true, ftox[child], xtof[t[child]]); } } else if (dep[a] == dep[b] + 1) { int ca = go_up(a, dep[a] - lca - 1); int cb = go_up(b, dep[b] - lca - 1); int meet = t[ca]; return g[meet][ftox[ca]].tot + calculate(meet, false, ftox[ca], ftox[cb]); } else if (dep[a] < dep[b]){ int temp = dep[b] - dep[a]; int mid = lca + (temp + 1)/ 2; int child = go_up(b, dep[b] - mid - 1); if (temp % 2){ return g[t[child]][xtof[t[child]]].tot + calculate(t[child], false, xtof[t[child]], ftox[child]); } else{ return g[t[child]][xtof[t[child]]].tot + calculate(t[child], true, xtof[t[child]], ftox[child]); } } else if(dep[a] == dep[b]) { int ca = go_up(a, dep[a] - lca - 1); int cb = go_up(b, dep[b] - lca - 1); int meet = t[ca]; return g[meet][ftox[ca]].tot + calculate(meet, true, ftox[ca], ftox[cb]); } assert(false); } void start() { for (int i = 1; i <= n; i++) { vis[i] = false; } t[0] = t[1] = -1; dfs(1); treedp(1); for (int now = 1; now <= n; now++) { sort(g[now].begin(), g[now].end()); for (int i =(int) g[now].size() - 1; i >= 0; i--) { g[now][i].ss = g[now][i].tot; if (i + 3 <= (int)g[now].size()) { g[now][i].ss += g[now][i + 2].ss; } } } df = 1; euler_circuit(1, 0); get_up_node(); sparse_table(); relation(); } int main() { for (int i = 0; i < maxlevel; i++){ if ( (1<<i) < maxn*2) lg2[1 << i] = i; } for (int i = 3; i < maxn*2; i++) { if (!lg2[i]){ lg2[i] = lg2[i - 1]; } } int T; cin >> T; while (T--) { all = 0; cin >> n >> m; for (int i = 0; i <= n; i++){ g[i].clear(); } Node node; for (int i = 1; i <= n - 1; i++) { //cin >> node.now >> node.to >> node.c; SS(node.now); SS(node.to); SS(node.c); g[node.now].push_back(node); swap(node.now, node.to); g[node.now].push_back(node); all += node.c; } start(); int a, b; for (int i = 1; i <= m; i++){ //cin >> a >> b; SS(a); SS(b); //cout << find(a, b) << endl; printf("%d\n", find(a, b)); } } return 0; }
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