codeforces605

A. Three Friends

#include<bits/stdc++.h>
#define rep(i, n) for(int i=0;i!=n;++i)
#define per(i, n) for(int i=n-1;i>=0;--i)
#define Rep(i, sta, n) for(int i=sta;i!=n;++i)
#define rep1(i, n) for(int i=1;i<=n;++i)
#define per1(i, n) for(int i=n;i>=1;--i)
#define Rep1(i, sta, n) for(int i=sta;i<=n;++i)
#define L k<<1
#define R k<<1|1
#define inf (0x3f3f3f3f)
#define llinf (1e18)
#define mid (tree[k].l+tree[k].r)>>1
#define ALL(A) A.begin(),A.end()
#define SIZE(A) ((int)A.size())
typedef long long i64;
using namespace std;
int main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int q,a,b,c;  cin >> q;
    while(q--){
        cin >> a >> b >> c;
        int minn = min(min(a,b),c);
        int maxx = max(max(a,b),c);
        cout << max(0,(maxx-minn)*2-4) <<'\n'; 
    }
    return 0;
}

B. Snow Walking Robot

#include<bits/stdc++.h>
#define rep(i, n) for(int i=0;i!=n;++i)
#define per(i, n) for(int i=n-1;i>=0;--i)
#define Rep(i, sta, n) for(int i=sta;i!=n;++i)
#define rep1(i, n) for(int i=1;i<=n;++i)
#define per1(i, n) for(int i=n;i>=1;--i)
#define Rep1(i, sta, n) for(int i=sta;i<=n;++i)
#define L k<<1
#define R k<<1|1
#define inf (0x3f3f3f3f)
#define llinf (1e18)
#define mid (tree[k].l+tree[k].r)>>1
#define ALL(A) A.begin(),A.end()
#define SIZE(A) ((int)A.size())
typedef long long i64;
using namespace std;
int main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int q;  cin >> q;
    while(q--){
        string str; cin >> str;
        int numU = 0,numD = 0,numL = 0,numR = 0;
        for(int i=0;i!=SIZE(str);++i)
            switch(str[i]){
                case 'D': ++numD;   break;
                case 'U': ++numU;   break;
                case 'L': ++numL;   break;
                case 'R': ++numR;   break;
            }
        int r = min(numD,numU),c = min(numL,numR);
        if(!r&&!c){
            cout << 0 << '\n';
            continue;
        }
        if(!c){
            cout << 2 <<'\n';
            cout << "UD" <<'\n';
            continue;
        }
        if(!r){
            cout << 2 <<'\n';
            cout << "LR" <<'\n';
            continue;
        }
        cout << (r + c)*2 <<'\n';
        for(int i=0;i!=r;++i)
            cout << "U" ;
        for(int i=0;i!=c;++i)
            cout << "L" ;
        for(int i=0;i!=r;++i)
            cout << "D" ;
        for(int i=0;i!=c;++i)
            cout << "R" ;
        cout << '\n';
    }    
    return 0;

C. Yet Another Broken Keyboard

#include<bits/stdc++.h>
#define rep(i, n) for(int i=0;i!=n;++i)
#define per(i, n) for(int i=n-1;i>=0;--i)
#define Rep(i, sta, n) for(int i=sta;i!=n;++i)
#define rep1(i, n) for(int i=1;i<=n;++i)
#define per1(i, n) for(int i=n;i>=1;--i)
#define Rep1(i, sta, n) for(int i=sta;i<=n;++i)
#define L k<<1
#define R k<<1|1
#define inf (0x3f3f3f3f)
#define llinf (1e18)
#define mid (tree[k].l+tree[k].r)>>1
#define ALL(A) A.begin(),A.end()
#define SIZE(A) ((int)A.size())
typedef long long i64;
using namespace std;
int main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int n,k;    cin >> n >> k;
    string str; cin >> str;
    set<char> S;    char ch;
    while(k--){
        cin >> ch;
        S.insert(ch);
    }
    i64 len = 0,sum = 0;
    bool flag = false;
    for(int i=0;i!=SIZE(str);++i){
        if(!flag&&S.count(str[i])){
            ++len;
            flag = true;
            continue;
        }
        if(flag){
            if(S.count(str[i]))
                ++len;
            else{
                sum += (1+len)*len/2;
                len = 0;
                flag = false;
            }
        }
    }
    sum += (1+len)*len/2;
    cout << sum <<'\n';
    return 0;
}

D. Remove One Element

题解：求出每一个位置往左和往右的最长位置，枚举每一个中间位置，如果左边小于右边则与最长长度比较

#include<bits/stdc++.h>
#define rep(i, n) for(int i=0;i!=n;++i)
#define per(i, n) for(int i=n-1;i>=0;--i)
#define Rep(i, sta, n) for(int i=sta;i!=n;++i)
#define rep1(i, n) for(int i=1;i<=n;++i)
#define per1(i, n) for(int i=n;i>=1;--i)
#define Rep1(i, sta, n) for(int i=sta;i<=n;++i)
#define L k<<1
#define R k<<1|1
#define inf (0x3f3f3f3f)
#define llinf (1e18)
#define mid (tree[k].l+tree[k].r)>>1
#define ALL(A) A.begin(),A.end()
#define SIZE(A) ((int)A.size())
typedef long long i64;
using namespace std;
int main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int n;  cin >> n;
    vector<int> arr(n),l(n),r(n);
    for(auto& v:arr)
        cin >> v;
    for(int i=0;i!=n;++i)
        l[i] = i>0&&arr[i]>arr[i-1] ? l[i] = l[i-1] + 1 : 1;
    for(int i=n-1;i>=0;--i)
        r[i] = i<n-1&&arr[i]<arr[i+1] ? r[i] = r[i+1] + 1 : 1;
    int maxValue = *max_element(l.begin(),l.end());
    for(int index=1;index!=n-1;++index)
        if(arr[index-1] < arr[index+1])
            maxValue = max(maxValue,l[index-1]+r[index+1]);
    cout << maxValue << '\n';
    return 0;
}

E. Nearest Opposite Parity

题解：将多源点转换为单点，因此反向建边寻找最短路径

#include<bits/stdc++.h>
#define rep(i, n) for(int i=0;i!=n;++i)
#define per(i, n) for(int i=n-1;i>=0;--i)
#define Rep(i, sta, n) for(int i=sta;i!=n;++i)
#define rep1(i, n) for(int i=1;i<=n;++i)
#define per1(i, n) for(int i=n;i>=1;--i)
#define Rep1(i, sta, n) for(int i=sta;i<=n;++i)
#define L k<<1
#define R k<<1|1
#define inf (0x3f3f3f3f)
#define llinf (1e18)
#define mid (tree[k].l+tree[k].r)>>1
#define ALL(A) A.begin(),A.end()
#define SIZE(A) ((int)A.size())
#define PII pair<int,int>
typedef long long i64;
using namespace std;
const int maxn = 2e5 + 32;
int arr[maxn],ans[maxn],dis[maxn];
vector<int> Edge[maxn];
void spfa(int cur){
    memset(dis,inf,sizeof(dis));
    queue<int> q;   q.push(cur);
    dis[cur] = 0;
    while(!q.empty()){
        int c = q.front();  q.pop();
        for(auto& v:Edge[c])
            if(dis[v] > dis[c] + 1){
                dis[v] = dis[c] + 1;
                q.push(v);
            }
    }
}
int main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int n;  cin >> n;
    for(int i=1;i<=n;++i){
        cin >> arr[i];
        if(i-arr[i] >= 1) Edge[i-arr[i]].push_back(i);
        if(i+arr[i] <= n) Edge[i+arr[i]].push_back(i);  
    }
    //0 odd n+1 even
    for(int i=1;i!=n+1;++i){
        if(arr[i]&1)    Edge[n+1].push_back(i);
        else    Edge[0].push_back(i);
    }
    memset(ans,-1,sizeof(ans));
    spfa(0);
    for(int i=1;i<=n;++i)
        if(arr[i]&1&&dis[i]!=inf)    ans[i] = dis[i];
    spfa(n+1);               
    for(int i=1;i<=n;++i)
        if(!(arr[i]&1)&&dis[i]!=inf)    ans[i] = dis[i];
    for(int i=1;i<=n;++i)
        cout << (ans[i] == -1 ? -1:ans[i] -1) <<" \n"[i==n];  
    return 0;
}

F题没写出来，菜得真实