2018 计蒜之道 初赛 第一场

A-百度无人车

题解：典型的二分

#pragma warning(disable:4996)
#include<map>
#include<queue>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long 
using namespace std;

const int maxn = 20005;

ll s;
int n, p;
int a[maxn];

bool check(int mid) {
    int pos = lower_bound(a + 1, a + n + 1, mid) - a;
    int ans = 0;
    for (int i = pos; i <= n; i++) ans += (a[i] - mid);
    if (ans > s / p) return false;
    return true;
}

int Solve() {
    int l = 1, r = a[n];
    for (int i = 1; i <= 30; i++) {
        int mid = (l + r) >> 1;
        if (check(mid)) r = mid;
        else l = mid + 1;
    }
    return l;
}

int main()
{
    while (scanf("%d", &n) != EOF) {
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        sort(a + 1, a + n + 1);
        scanf("%d%lld", &p, &s);
        printf("%d
", Solve());
    }
    return 0;
}

B-百度科学家

题解：数据比较小的时候建图，跑DFS统计总的污染值即可。

感受：题意不明！！！！！书替换后不会对以前的结果造成影响。所以~~要n何用。

#pragma warning(disable:4996)
#include<vector>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long 
using namespace std;

const ll INF = 1000000000000000;

int n, m;
int p[30], pos[30];

vector<int> book[30];

bool use[30];

ll dp[30];

void DFS(int u) {
    use[u] = 1;
    dp[u] = p[u];
    for (int i = 0; i < book[u].size(); i++) {
        int v = book[u][i];
        if (!use[v]) {
            DFS(v);
            dp[u] += dp[v];
        }
    }
}

int main()
{
    while (cin >> n) {
        for (int i = 1; i <= n; i++) pos[i] = i;
        for (int i = 1; i <= n; i++) cin >> p[i];

        cin >> m;
        int k = n;
        for (int i = 1; i <= m; i++) {
            int op;
            cin >> op;
            if (op == 0) {
                int x, v;
                cin >> x >> v;
                pos[x] = (++k);
                p[k] = v;
            }
            else {
                int x, l, r;
                cin >> x >> l >> r;
                for (int i = l; i <= r; i++) book[pos[x]].push_back(pos[i]);
            }
        }

        ll ans = INF;
        for (int i = 1; i <= k; i++) {
            memset(use, 0, sizeof(use));
            memset(dp, 0, sizeof(dp));
            DFS(i);
            ans = min(ans, dp[i]);
        }
        cout << ans << endl;

        for (int i = 1; i <= k; i++) book[i].clear();
    }
    return 0;
}