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2018 计蒜之道 初赛 第一场

时间:2018-05-12 21:38:15      阅读:258      评论:0      收藏:0      [点我收藏+]

标签:class   return   cloc   open   oid   scanf   百度   div   ide   

百度无人车

二分

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#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 2e4 + 10;
LL a[maxn];

int main(){
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) scanf("%lld", a + i);
    LL p, s;
    scanf("%lld %lld", &p, &s);
    int l = 1, r = 20000;
    while(l < r){
        int mid = (l + r) >> 1;
        LL t = 0;
        for(int i = 1; i <= n; ++i) t += max(0LL, a[i] - mid);
        if(t * p <= s) r = mid;
        else l = mid + 1;
    }
    printf("%d\n", r);
    return 0;
}
Aguin

 百度科学家

主席树建图

技术分享图片
  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 typedef long long LL;
  4 const int maxn = 1e5 + 10;
  5 int a[maxn];
  6 
  7 LL f[maxn * 35];
  8 
  9 // Persistent SegmentTree
 10 vector<int> G[maxn * 35];
 11 
 12 struct node {
 13     int tag, ls, rs;
 14 } nod[maxn * 35];
 15 
 16 int cnt;
 17 int id[maxn];
 18 void build(int p, int tl, int tr){
 19     cnt++;
 20     if(tl == tr){
 21         id[tl] = p;
 22         nod[p].tag = a[tl];
 23         return;
 24     }
 25     int mid = (tl + tr) >> 1;
 26     nod[p].ls = cnt + 1, G[p].push_back(cnt + 1), build(cnt + 1, tl, mid);
 27     nod[p].rs = cnt + 1, G[p].push_back(cnt + 1), build(cnt + 1, mid + 1, tr);
 28 }
 29 void update(int l, int r, int &x, int y, int pos, int v) {
 30     nod[++cnt] = nod[y];
 31     x = cnt;
 32     if (l == r) {
 33         id[l] = x;
 34         nod[x].tag = v;
 35         return;
 36     }
 37     int mid = (l + r) >> 1;
 38     if (pos <= mid) update(l, mid, nod[x].ls, nod[y].ls, pos, v);
 39     else update(mid + 1, r, nod[x].rs, nod[y].rs, pos, v);
 40     G[x].push_back(nod[x].ls);
 41     G[x].push_back(nod[x].rs);
 42 }
 43 void query(int p, int tl, int tr, int l, int r, int x)
 44 {
 45     if(tr < l || r < tl) return;
 46     if(l <= tl && tr <= r){
 47         G[x].push_back(p);
 48         return;
 49     }
 50     int mid = (tl + tr) >> 1;
 51     query(nod[p].ls, tl, mid, l, r, x);
 52     query(nod[p].rs, mid + 1, tr, l, r, x);
 53     return;
 54 }
 55 
 56 // Tarjan
 57 stack<int> S;
 58 int dfs_clock, dfn[maxn * 35], low[maxn * 35];
 59 int bcc_cnt, bccno[maxn * 35];
 60 void dfs(int u)
 61 {
 62     dfn[u] = low[u] = ++dfs_clock;
 63     S.push(u);
 64     for(int i = 0; i < G[u].size(); i++)
 65     {
 66         int v = G[u][i];
 67         if(!dfn[v])
 68         {
 69             dfs(v);
 70             low[u] = min(low[u], low[v]);
 71         }
 72         else if(!bccno[v]) low[u] = min(low[u], dfn[v]);
 73     }
 74 
 75     if(low[u] == dfn[u])
 76     {
 77         bcc_cnt++;
 78         while(1)
 79         {
 80             int x = S.top(); S.pop();
 81             bccno[x] = bcc_cnt;
 82             if(x == u) break;
 83         }
 84     }
 85 }
 86 
 87 void find_bcc(int n)
 88 {
 89     memset(dfn, 0, sizeof(dfn));
 90     memset(bccno, 0, sizeof(bccno));
 91     dfs_clock = bcc_cnt = 0;
 92     for(int i = 1; i <= n; i++) if(!dfn[i]) dfs(i);
 93 }
 94 
 95 int main(){
 96     int N, M, rt = 1;
 97     scanf("%d", &N);
 98     for(int i = 1; i <= N; ++i) scanf("%d", a + i);
 99     build(1, 1, N);
100     scanf("%d", &M);
101     while(M--){
102         int op, x, y, l, r;
103         scanf("%d %d", &op, &x);
104         if(op == 0){
105             scanf("%d", &y);
106             update(1, N, rt, rt, x, y);
107         }
108         else{
109             scanf("%d %d", &l, &r);
110             query(rt, 1, N, l, r, id[x]);
111         }
112     }
113     assert(cnt < maxn * 35);
114     find_bcc(cnt);
115     for(int i = 1; i <= cnt; ++i){
116         f[bccno[i]] += nod[i].tag;
117         for(int j = 0; j < G[i].size(); ++j){
118             int to = G[i][j];
119             if(bccno[to] != bccno[i]) f[bccno[i]] = 1e18;
120         }
121     }
122     LL ans = 1e18;
123     for(int i = 1; i <= bcc_cnt; ++i){
124         ans = min(ans, f[i]);
125     }
126     printf("%lld\n", ans);
127     return 0;
128 }
Aguin

 

2018 计蒜之道 初赛 第一场

标签:class   return   cloc   open   oid   scanf   百度   div   ide   

原文地址:https://www.cnblogs.com/Aguin/p/9029943.html

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