HDU 4714 Tree2cycle（树状DP）（2013 ACM/ICPC Asia Regional Online ―― Warmup）

Description

A tree with N nodes and N-1 edges is given. To connect or disconnect one edge, we need 1 unit of cost respectively. The nodes are labeled from 1 to N. Your job is to transform the tree to a cycle(without superfluous edges) using minimal cost. 
A cycle of n nodes is defined as follows: (1)a graph with n nodes and n edges (2)the degree of every node is 2 (3) each node can reach every other node with these N edges.

 

Input

The first line contains the number of test cases T( T<=10 ). Following lines are the scenarios of each test case.  In the first line of each test case, there is a single integer N( 3<=N<=1000000 ) - the number of nodes in the tree. The following N-1 lines describe the N-1 edges of the tree. Each line has a pair of integer U, V ( 1<=U,V<=N ), describing a bidirectional edge (U, V). 

 

Output

For each test case, please output one integer representing minimal cost to transform the tree to a cycle. 

题目大意：一棵有n个点的树，删边需要1的费用，增边需要1的费用，问最少需要多少费用才能得到一个环，不能用多余的边（即总共n条边）。

思路：首先我们可以这样考虑：我们先删掉x条边，那么之后再加上x+1条边，形成一个环。我们删掉x条边后，所有的点的度都不能大于2，那么就会出现多条链，再用x+1条边把这些链首尾相接就可以形成一个环。现在问题就转化成了给一棵树，问最少删掉多少条边，使得每个点的度不大于2。然后就是树状DP，用dp[i][0]表示，第i个点，连0个或1个子节点（度小于2）的最小费用。用dp[i][1]表示，第i个点，连0个或1个或2个子节点（度小于等于2）的最小费用。这样对每个点选择是不连或者连一个子节点，还是连两个子节点。然后随便搞，时间复杂度为O(n)。

PS：100W个点我看到好多人栈溢出了所以大家还是写非递归吧（实际上会不会溢出我不知道我没试过我一开始就写非递归）……我极少写非递归可能写得比较挫……

代码（2703MS）：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int MAXN = 1000010;
const int MAXE = MAXN * 2;

int head[MAXN];
int stk[MAXN], stkp[MAXN], top;
int next[MAXE], to[MAXE];
int ecnt, n, T;

void init() {
    memset(head, 0, sizeof(head));
    ecnt = 1;
}

void add_edge(int u, int v) {
    to[ecnt] = v; next[ecnt] = head[u]; head[u] = ecnt++;
    to[ecnt] = u; next[ecnt] = head[v]; head[v] = ecnt++;
}

int dp[MAXN][2];
//0:连0 or 1个子节点，1:连两个子节点
int solve() {
    top = 1;
    stk[top] = 1; stkp[top] = head[1];
    while(top > 0) {
        int &p = stkp[top], u = stk[top];
        if(to[p] == stk[top - 1]) p = next[p];
        if(p) {
            int &v = to[p];
            ++top; stk[top] = v; stkp[top] = head[v];
            p = next[p];
        }
        else {
            int min1 = MAXN, min2 = MAXN;
            dp[u][0] = 0;
            for(int q = head[u]; q; q = next[q]) {
                int &v = to[q];
                if(v == stk[top - 1]) continue;
                ++dp[u][0];
                dp[u][0] += min(dp[v][0], dp[v][1]);
                int x = dp[v][0] - min(dp[v][0], dp[v][1]);
                if(x < min1) {
                    min2 = min1;
                    min1 = x;
                }
                else min2 = min(min2, x);
            }
            int best = dp[u][0];
            dp[u][0] = min(dp[u][0], best - 1 + min1);
            dp[u][1] = min(dp[u][0], best - 2 + min1 + min2);
            --top;
        }
    }
    return 2 * min(dp[1][0], dp[1][1]) + 1;
}

int main() {
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        init();
        int u, v;
        for(int i = 1; i < n; ++i) {
            scanf("%d%d", &u, &v);
            add_edge(u, v);
        }
        printf("%d
", solve());
    }
}