POJ3659 Cell Phone Network

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<queue>#include<cmath>#include<memory.h>using namespace std;#define inf (1 << 29)#define Min(a, b) (a) < (b) ? (a) : (b)#define maxn 21000struct T{int v, next;}fn[maxn];int th;int g[maxn], dp[maxn][3], degree[maxn];void add(int u, int v){fn[th].v = v, fn[th].next = g[u], g[u] =th++;}void dfs(int u, int f){if (u != 1 && degree[u] == 1){dp[u][0] = 1;dp[u][1] = inf;dp[u][2] = 0;return;}dp[u][0] = 1;int i, v, sign, min, del;for (i = g[u], sign = 0, min = inf; i != -1; i = fn[i].next){v = fn[i].v;if (v == f){continue;}dfs(v, u);    //如果取u，则孩子v取MIN(的可取[v被自己覆盖]，依赖孩子的孩子[表示v不用已经被覆盖，不依赖u]，依赖父节点u[因为父节点现在取了])dp[u][0] += Min(Min(dp[v][0], dp[v][1]), dp[v][2]);if (dp[v][0] <= dp[v][1])//u不取也不依赖于父节点，则u依赖于孩子，保证一个孩子被取。{dp[u][1] += dp[v][0];sign = 1;//选了一个有电话的，标志1}else//选了一个孩子没电话的，要保存min，如果sign没标志过，就得利用有电话的最小值得那个人。{dp[u][1] += dp[v][1];if (min > dp[v][0]){min = dp[v][0];del = dp[v][1];}}dp[u][2] += Min(dp[v][0], dp[v][1]);//u不取，则u依赖于父节点，则取v不依赖与u的最优情况}if (!sign){dp[u][1] += min - del;}}int main(){int n, u, v, i, j;while (scanf("%d", &n)==1){th = 0;for (i = 0; i <= n; i++){degree[i] = 0;g[i] = -1;dp[i][0] = dp[i][1] = dp[i][2] = 0;}for (i = 1; i < n; i++){scanf("%d%d", &u, &v);add(u, v);add(v, u);degree[u]++;degree[v]++;}if (n == 1){printf("0\n");continue;}dfs(1, 0);printf("%d\n", Min(dp[1][0], dp[1][1]));}return 0;}