牛客IOI周赛18-普及组(A-C)

A、数字计数
显然水题

def main():
    n = int(input())
    a_li = list(set(map(int,input().split())))
    a_li.sort()
    print(a_li[-1] - a_li[-2],a_li[-1] - a_li[1],a_li[-2] - a_li[1],a_li[-2] - a_li[0])

if __name__ == "__main__":
    main()

B、数颜色
一开始想暴力，想了想可能会超时，然后想到了为什么不枚举颜色，求出可能出现的区间个数呢,而我写的算法是 颜色个数*n*(n+1)/2 - 某颜色不可能出现的区间个数

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int NMAX = 2e5 + 10;
const int MOD = 1e9 + 7;
int pre[NMAX];
int main(int argc, char const *argv[])
{
    int n , a;
    scanf("%d",&n);
    ll ans = 0;
    for(register int i = 1;i <= n;i++)
    {
        scanf("%d",&a);
        if(!pre[a]) ans += n * (n + 1) / 2;
        ans -= (i - pre[a] - 1) * (i - pre[a]) / 2;
        pre[a] = i;
    }
    for(register int i = 1;i <= 1000;i++)
    {
        if(!pre[i])    continue;
        ans -= (n + 1 - pre[i] - 1) * (n + 1 - pre[i]) / 2;
    }
    printf("%lld
",ans );
    return 0;
}

C、智斗恶龙

bfs, 但vised不能每次都全部进行清零，只对修改过的vised数组进行清零

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int NMAX = 1e3 + 10;
const int MOD = 1e9 + 7;

ll mmap[NMAX][NMAX] , gold[NMAX*NMAX];
bool vised[NMAX][NMAX];
pair<int,int>vec[NMAX*NMAX];
struct Node{
    int x,y,step;
    Node(int x,int y,int step):x(x),y(y),step(step){};
};
int tot = 0 , cot = 0;
int dir[4][2] = {-1,0,1,0,0,1,0,-1};

int main(int argc, char const *argv[])
{
    int t;
    int n , m , sx , sy , d , x;
    scanf("%d",&t);
    set<ll>s;
    while(t--)
    {
        s.clear();
        cot = tot = 0;
        scanf("%d%d%d%d%d%d",&n,&m,&sx,&sy,&d,&x);
        for(register int i = 1;i <= n;i++)
            for(register int j = 1;j <= m;j++)
                scanf("%lld",&mmap[i][j]);
        if(mmap[sx][sy] == -1)
        {
            puts("no");
            continue;
        }
        queue<Node>q;
        if(mmap[sx][sy])
            s.insert(mmap[sx][sy]);
        q.push(Node(sx,sy,0));
        vised[sx][sy] = true;
        vec[tot++] = make_pair(sx,sy);
        while(!q.empty())
        {
            Node now = q.front();q.pop();
            for(register int i = 0;i < 4;i++)
            {
                int next_x , next_y;
                next_x = now.x + dir[i][0];
                next_y = now.y + dir[i][1];
                if(next_x < 1 || next_x > n || next_y < 1 || next_y > m || mmap[next_x][next_y] == -1|| vised[next_x][next_y] || now.step >= d)    continue;
                vec[tot++] = make_pair(next_x,next_y);
                vised[next_x][next_y] = true;
                q.push(Node(next_x,next_y,now.step + 1));
                if(mmap[next_x][next_y])
                    s.insert(mmap[next_x][next_y]);
            }
        }
        while(tot)
        {
            int next_x = vec[tot-1].first , next_y = vec[tot-1].second;
            vised[next_x][next_y] = false;
            tot--;
        }
        if(s.size() < x)
            puts("no");
        else
        {
            for(set<ll>::iterator it = s.begin(); it != s.end();it++)
                gold[++cot] = *it;
            ll ans = gold[cot] - gold[1];
            for(register int i = 1;i <= cot - x + 1;i++)
                ans = min(ans , gold[i + x - 1] - gold[i]);
            printf("%lld
",ans );
        }
    }
    return 0;
}