AOJ 2456 Usoperanto （树形dp）

题意:

N<=106单词,给出每个单词的单词字母个数,并给出N个单词之间的修饰关系
任意两个修饰单词与被修饰单词的花费 是中间夹着的所有单词的长度和
求如何安排单词之间的顺序, 使得任意两个修饰单词与被修饰单词花费的总和最小, 输出这个总和

分析:

对于多个修饰词同时修饰一个被修饰词的情况, 我们自然而然的考虑到一个贪心, 越排在外面的被算的次数越少, 显然从大到小排序计算.
问题来了 如果有嵌套多重修饰的情况 a−>b,bcf−>g 且 b 的大小是位于中间的, 那么之前的贪心就失效了.
那该怎么办呢 由于修饰关系构成了一棵树, 对于某一颗子树的修饰关系我们肯定是要一起考虑的, 不然就对别的子树有了更多的贡献, 这样结果是不优的.
考虑到这样,我们就可以树形dp了,dp[i]:=以i为根的子树修饰关系被处理完的最小花费
那么转移dp[u]=∑sonsi=0dp[v]+∑sonsi=0i∗sum[v](sum[v] is in decreasing order)

坑:

会爆栈 需要手动模拟递归 或者 STL stack bfs 模拟也可以

代码:
模拟递归

//
// Created by TaoSama on 2015-10-02
// Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e6 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

typedef long long LL;

int fa[N], cur[N], head[N], nxt[N], pnt[N], cnt;
void add_edge(int u, int v) {
    pnt[cnt] = v;
    nxt[cnt] = head[u];
    head[u] = cnt++;
}

int stk[N], top;

int n, val[N], sum[N], a[N];
LL ans;

void calc(int u) {
    int sz = 0;
    for(int i = head[u]; ~i; i = nxt[i]) {
        int v = pnt[i];
        a[sz++] = sum[v];
    }
    sort(a, a + sz, greater<int>());
    sum[u] = val[u];
    for(int i = 0; i < sz; ++i) {
        sum[u] += a[i];
        ans += 1LL * i * a[i];
    }
}

void dfs(int s) {
    top = 0;
    stk[++top] = s;
    while(top) {
        int u = stk[top];
        if(~cur[u]) {
            int &i = cur[u];
            int v = pnt[i];
            stk[++top] = v;
            i = nxt[i];
        } else {
            calc(u);
            --top;
        }
    }
}

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d", &n) == 1) {
        cnt = 0; memset(head, -1, sizeof head);
        for(int v = 0; v < n; ++v) {
            scanf("%d%d", val + v, fa + v);
            if(~fa[v]) add_edge(fa[v], v);
        }
        ans = 0;
        memcpy(cur, head, sizeof head);
        for(int i = 0; i < n; ++i) {
            if(fa[i] == -1) dfs(i);
        }
        printf("%lld\n", ans);
    }
    return 0;
}

STL stack bfs 模拟

//
// Created by TaoSama on 2015-10-02
// Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <stack>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e6 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

vector<int> G[N];

typedef long long LL;

int n, fa[N], sum[N];
LL dp[N];

bool cmp(int x, int y) {
    return sum[x] > sum[y];
}

void bfs(int root) {
    queue<int> q;
    stack<int> s;
    q.push(root); s.push(root);
    while(q.size()) {
        int u = q.front(); q.pop();
        for(int i = 0; i < G[u].size(); ++i) {
            int v = G[u][i];
            q.push(v);
            s.push(v);
        }
    }
    while(s.size()) {
        int u = s.top(); s.pop();
        dp[u] = 0;
        sort(G[u].begin(), G[u].end(), cmp);
        for(int i = 0; i < G[u].size(); ++i) {
            int v = G[u][i];
            dp[u] += dp[v] + 1LL * i * sum[v];
            sum[u] += sum[v];
        }
    }
}

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d", &n) == 1) {
        for(int i = 0; i < n; ++i) G[i].clear();
        for(int v = 0; v < n; ++v) {
            scanf("%d%d", sum + v, fa + v);
            if(~fa[v]) G[fa[v]].push_back(v);
        }
        LL ans = 0;
        for(int i = 0; i < n; ++i) {
            if(fa[i] == -1) {
                bfs(i);
                ans += dp[i];
            }
        }
        printf("%lld\n", ans);
    }
    return 0;
}