Codeforces Round #408 (Div. 2)ABCD

A. Buying A House

分别向左右枚举，找到符合题意的点，输出最短的长度*10。

#include <bits/stdc++.h>

using namespace std;

int n, k, s, a[105];
int main()
{
    scanf("%d%d%d", &n, &s, &k);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    int m = 100;
    for(int i = s+1; i <= n; i++){
        if(a[i] > 0 && a[i] <= k) {
            m = i-s;
            break;
        }
    }
    for(int i = s-1; i > 0; i--){
        if(a[i] > 0&& a[i] <= k){
            m = min(s-i, m);
            break;
        }
    }
    printf("%d\n", m*10);
    return 0;
}

B. Find The Bone

模拟即可，如果掉进洞里输出，否则输出最后结果

#include <bits/stdc++.h>

using namespace std;

int n, m, k, a[1000005];
int main() {
    scanf("%d%d%d", &n, &m, &k);

    for(int i = 1, x; i <= m; i++) {
        scanf("%d", &x);
        a[x] = 1;
    }
    if(a[1]) return 0*printf("1\n");
    else {
        int u, v, x = 1;
        while(k--) {
            scanf("%d%d", &u, &v);
            if(x == u) {
                x = v;
                if(a[v]) return 0*printf("%d\n", x);
            } else if(x == v) {
                x = u;
                if(a[u]) return 0*printf("%d\n", x);
            }
        } printf("%d\n", x);
    }

    return 0;
}

C. Bank Hacking

根据题意可知最大的结果不会超过 max{a_i}+2，

设寻找的第一个点为x, x周围的点为第二次寻找的点为y，设z为x外圈的点，寻找的第一点对答案的贡献为 a[x]

y对答案贡献为a[y]+1, z对答案恭喜为a[z]+2,因此决定答案的只有最大值max{a_i}和次大值{max_ai}-1（该值也可能不存在）

因此我们可以枚举每个点作为第一次寻找的点，并记录该点最大值和次大值对最后答案的贡献

#include <bits/stdc++.h>

using namespace std;

int n, m, k, s, a[300005];

int vis_small[300005], vis_big[300005];
map<int, int>Map;

int main() {
    scanf("%d", &n);
    s = 1;
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        Map[a[i]]++;
        if(a[i] > a[s]) s = i;
    }
    int big = a[s], small = big - 1, big_num = Map[big], small_num = Map[small];
    //printf("%d %d %d %d\n", big, small, big_num, small_num);

    for(int i = 0, u, v; i < n-1; i++) {
        scanf("%d%d", &u, &v);
        if(a[u] == big) vis_big[v]++;
        else if(a[u] == small) vis_small[v]++;
        if(a[v] == big) vis_big[u]++;
        else if(a[v] == small) vis_small[u]++;
    }
    int ans = big+2;
    for(int i = 1; i <= n; i++) {
        int xx, yy;
        if(a[i] == big) {
            if(big_num == 1) xx = big;
            else if(vis_big[i] == big_num-1) {
                xx = big+1;
            } else xx = big+2;
            if(small_num == 0) yy = xx;
            else if(vis_small[i] == small_num) {
                yy = big;
            } else yy = big+1;
        } else if(a[i] == small) {
            if(vis_big[i] == big_num) {
                xx = big+1;
            } else xx = big+2;
            yy = xx;
        } else {
            if(vis_big[i] == big_num) {
                xx = big+1;
            } else xx = big+2;
            yy = xx;
        }
        ans = min(ans, max(xx, yy));
    }
    printf("%d\n", ans);

    return 0;
}

D. Police Stations

把所有的police同时放入队列，然后bfs.用map记录需要连接的边

#include <bits/stdc++.h>

using namespace std;

int n, k, d, cnt = 0;
int vis[300005], x[300005], y[300005];

typedef pair<int, int> P;
vector<int>G[300005];
queue<int>que;
map<P, int>Map;

int main() {
    scanf("%d%d%d", &n, &k, &d);
    for(int i = 1, x; i <= k; i++) {
        scanf("%d", &x);
        que.push(x);
        vis[x] = 1;
    }
    for(int i = 0, u, v; i < n-1; i++) {
        scanf("%d%d", &u, &v);
        x[i] = u, y[i] = v;
        Map[P(u, v)] = 1;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    while(!que.empty()) {
        int u = que.front();
        que.pop();
        for(auto v : G[u]) {
            if(!vis[v]) {
                que.push(v);
                Map[P(u, v)] = 0;
                Map[P(v, u)] = 0;
                cnt++;
                vis[v] = 1;
            }
        }
    }
    printf("%d\n", n-1-cnt);
    for(int i = 0; i < n-1; i++) {
        if(Map[P(x[i], y[i])] > 0) printf("%d ", i+1);
    }
    //system("pause");
    return 0;
}