UVa #1218 Perfect Service (例题9-14)

发现动态规划的时间复杂度等于dp数组开的大小。像这道题是n*3，复杂度为O(n)。之前有些题目是n*n，复杂度就是O(n^2)

不像例题9-13，这道题本身是无根树，记忆化搜索的函数要加一个父亲节点作为参数。

Run Time: 0.045s

#define UVa  "LT9-14.1218.cpp"		//Perfect Service
char fileIn[30] = UVa, fileOut[30] = UVa;

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>

using namespace std;

//Global Variables. Reset upon Each Case!
const int maxn = 10000 + 10, INF = maxn;
int N;
vector<int> S[maxn];
int d[maxn][3];
/////

int dp(int u, int s, int fa) {
    if(d[u][s] != -1) return d[u][s];
    if(S[u].size() == 1 && S[u][0] == fa){
        if(s == 0)      d[u][s] = 1;
        else if(s == 1) d[u][s] = 0;
        else if(s == 2) d[u][s] = INF;
        return d[u][s];
    }
    else {
        if(s == 0) {        //u is server.
            d[u][s] = 1;
            for(int i = 0; i < S[u].size(); i ++) {
                if(S[u][i] != fa)
                    d[u][s] += min(dp(S[u][i], 0, u), dp(S[u][i], 1, u));
            }
        }
        else if(s == 1) {   //fa is server, u is not.
            d[u][s] = 0;
            for(int i = 0; i < S[u].size(); i ++) {
                if(S[u][i] != fa)
                    d[u][s] += dp(S[u][i], 2, u);
            }
        }
        else if (s == 2) {  //neither fa or u is server.
            d[u][s] = INF;
            for(int i = 0; i < S[u].size(); i ++) {
                if(S[u][i] != fa)
                    d[u][s] = min(d[u][s], dp(u, 1, fa) - dp(S[u][i], 2, u) + dp(S[u][i], 0, u));
            }
        }
        return d[u][s];
    }
}

int main() {
    int END;
    do {
        for(int i = 0; i < maxn; i ++) S[i].clear();
        memset(d, -1, sizeof(d));

        scanf("%d", &N);
        int a, b;
        for(int i = 0; i < N-1; i ++) {
            scanf("%d%d", &a, &b);
            S[a].push_back(b);
            S[b].push_back(a);
        }
        printf("%d\n", min(dp(1,0,1),dp(1,2,1)));

    }while(scanf("%d", &END) && !END);

    return 0;
}