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  • 2018年浙江理工大学程序设计竞赛校赛 Problem I: 沙僧

    沙僧

    思路:

    dfs序+差分数组

    分层考虑,通过dfs序来查找修改的区间段,然后用差分数组修改

    代码:

    #include<bits/stdc++.h>
    using namespace std;
    #define fi first
    #define se second
    #define pi acos(-1.0)
    #define LL long long
    #define mp make_pair
    #define pb push_back
    #define ls rt<<1, l, m
    #define rs rt<<1|1, m+1, r
    #define ULL unsigned LL
    #define pll pair<LL, LL>
    #define pii pair<int, int>
    #define piii pair<int,pii>
    #define mem(a, b) memset(a, b, sizeof(a))
    #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
    //head
    
    const int N = 1e5 + 5;
    int deep[N], l[N], r[N];
    LL ans[N];
    vector<int>g[N];
    vector<int>d[N];
    vector<LL>v[N];
    vector<int>t[N];
    int x = 0;
    void dfs(int o, int u, int dd) {
        deep[u] = dd;
        d[dd].pb(x);
        t[dd].pb(u);
        v[dd].pb(0);
        l[u] = x;
        for (int i = 0; i < g[u].size(); i++) {
            if(g[u][i] != o) {
                ++x;
                dfs(u, g[u][i], dd+1);
            }
        }
        r[u] = x;
    }
    int main() {
        int n, m, a, b, c;
        scanf("%d %d", &n, &m);
        for (int i = 1; i < n; i++) {
            scanf("%d %d", &a, &b);
            g[a].pb(b);
            g[b].pb(a);
        }
        ++x;
        dfs(1, 1, 0);
        while(m --) {
            scanf("%d %d %d", &a, &b, &c);
            int pos = deep[a] + b;
            if(pos >= N || d[pos].size() == 0) continue;
            int st = lower_bound(d[pos].begin(), d[pos].end(), l[a]) - d[pos].begin();
            int ed = upper_bound(d[pos].begin(), d[pos].end(), r[a]) - d[pos].begin();
            v[pos][st] += c;
            if(ed < v[pos].size()) v[pos][ed] -= c;
        }
        for (int i = 0; i <= n; i++) {
            if(v[i].size() == 0) break;
            ans[t[i][0]] = v[i][0];
            for (int j = 1; j < v[i].size(); j++) {
                v[i][j] += v[i][j-1];
                ans[t[i][j]] = v[i][j];
            }
        }
        for (int i = 1; i <= n; i++) printf("%lld%c", ans[i], " 
    "[i==n]);
        return 0;
    }
    
    /*
    7 2
    1 2
    1 3
    2 4
    2 5
    3 6
    3 7
    2 1 2
    3 1 2
    */
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  • 原文地址:https://www.cnblogs.com/widsom/p/9279251.html
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