codeforces#600(div2)

A - Single Push

　　　题面意思，在给a数组取l - r加上k， 当a[i] - b[i] > 0 时，不成立，当出现0 - x - 0 - x的情况成立，当出现0 - x - y -0的情况不成立。那么其实只要在边界加上0，问题就变成了0 -x- 0活0 -0-0情况。

#include <cstdio>
 
int main(){
    int t, n, a[100002], b[100002], c[100002];
 
    scanf("%d", &t);
 
    for(int i = 0; i < t; i++){
        scanf("%d", &n);
 
        for(int j = 0; j < n; j++){
            scanf("%d", &a[j]);
        }
        for(int j = 0; j < n; j++){
            scanf("%d", &b[j]);
        }
        
        for(int j = 0; j < n; j++){
            c[j+1] = b[j] - a[j];
        }
        c[0] = 0;
        c[n + 1] = 0;
 
        int count = 0;
        for(int j = 0; j <= n; j++){
            if(c[j] < 0){
                count = 3;
                break;
            }
 
            if(c[j] != c[j + 1])
                count++;
        }
        
 
        if(count <= 2){
            printf("YES
");
        }
        else{
            printf("NO
");
        }
    }
 
}

B - Silly Mistake

　　　题面意思，把一个区间分别k区间，k个区间内满足，出现+x 必须出现-x，且同样的x不能出现在一个区间内出现俩次。

#include <bits/stdc++.h>
 
using namespace std;
 
const int N = 1e5 + 10;
 
int n;
int T;
int a[N];
set<int> q;
vector<int> ans;
map<int, int> ma;
 
int main()
{
 
    
        ma.clear();
        ans.clear();
        q.clear();
        scanf("%d", &n);
        
        for (int i = 1; i <= n; i ++)
        scanf("%d", &a[i]);
    //    cout << a[1] << endl;
        bool flag = true;
        for (int i = 1; i <= n; i ++)
        {
        //    cout << a[i] << endl;
            if(a[i] < 0)
            {
        //        cout << a[i] << endl;
                if(q.find(-a[i]) == q.end())
                {
                    flag = false;
                    break;
                }
                else
                {
                    q.erase(-a[i]);
                //    cout << q.size() << endl;
                    if(q.size() == 0)
                    {
                        ans.push_back(i);
                        ma.clear();
                    }
                }
            }
            else 
            {
                if(ma[a[i]])
                {
                    flag = false;
                    break;
                }
                else if(!ma[a[i]])
                {
                    q.insert(a[i]);
                    ma[a[i]] = 1;
                }
            }
        }
        if(q.size() > 0)
        {
            flag = false;
        }
        if(!flag )
        {
            puts("-1");
            return 0;
        }
        cout << ans.size() << endl;
        for (int i = 0; i < ans.size(); i ++)
        {
            if(i == 0)
            {
                cout << ans[i] << " ";
            }
            else
            {
                cout << ans[i] - ans[i - 1] << " ";
            }
        }
    
}

C - Sweets Eating

　　　　题面意思，吃糖, 每天最多吃m个糖，那么吃糖会有一个值d * a[i]， d为第几天吃，让你求吃1~n个糖的值得最小值。那么对于k个糖来说，a[i]越大的糖必须越先吃，才能使值最小，对于前i个糖， 那么它比前i - 1吃的糖的值，多出a[i] + a[i - m] + a[i - 2 *m] .... a[i - k * m],用前缀和来维护。

#include <bits/stdc++.h>
 
using namespace std;
 
const int N = 2e5 + 10;
typedef long long ll;
 
ll n, m;
ll a[N];
ll sum[N];
ll ans[N];
 
int main()
{
    scanf("%lld%lld", &n, &m);
    for (int i = 1; i <= n; i ++)
    {
        scanf("%lld", &a[i]);
    }
    sort(a + 1, a + n + 1);
    for (int i = 1; i <= m; i ++)
     sum[i] = a[i];
    for (int i = m; i <= n; i ++)
    {
        sum[i] = sum[i - m] + a[i];
    }
    for (int i = 1; i <= n; i ++)
    {
        ans[i] = ans[i - 1];
        ans[i] += sum[i];
    }
    for (int i = 1; i <= n; i ++)
    printf("%lld ", ans[i]);
    
}

D - Harmonious Graph

　　　　体面意思，给你一个无向图，然后对于任意一个联通块的最大点数r和最小的点数l，都存在r - l + 1 == 联通块的个数，所以可以先跑个tarjan求出联通块的数量，并记录联通块的最大值和最小值，根据左端点排序，那么现在来合并联通块， 对于要合并的联通块，一定存在交集，Li 一顶小于Ri + 1，此时把联通块合在一个更新R。

#include <bits/stdc++.h>
 
using namespace std;
 
const int N = 3e5 + 10;
const int M = 6e5 + 10;
const int INF = 0x3f3f3f3f;
 
 
 
int n, m;
int h[N], e[M], ne[M], idx;
int dfn[N], low[N], stac[N], num;
int top, st[N];
int cnt;
int c[N];
int p[N]; 
set<int> s[N];
void add(int a, int b)
{
    e[idx] = b;
    ne[idx] = h[a];
    h[a] = idx ++;
}
void tarjan(int u)
{
    dfn[u] = low[u] = ++ num;
    stac[++ top] = u;
    st[u] = 1;
    for (int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if(!dfn[j])
        {
            tarjan(j);
            low[u] = min(low[u], low[j]);
        }
        else if(st[j])
            low[u] = min(low[u], dfn[j]);
    }
    if(dfn[u] == low[u])
    {
        ++ cnt;
        int z;
        do{
            z = stac[top --];
            st[z] = 0;
            c[z] = cnt;
            p[z] = u;
             s[cnt].insert(z);
        }while(z != u);
    }
}
 
int find(int x)
{
    if(x == p[x]) return x;
    return p[x] = find(p[x]); 
} 
 
 
int main()
{
    memset(h, -1, sizeof(h));
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; i ++)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        add(a, b), add(b, a);
    }
 
    for (int i = 1; i <= n; i ++)
    {
        if(!dfn[i])
        tarjan(i);
    }
    for (int i = 1; i <= cnt; i ++)
     {
         p[i] = i;
     } 
    
    int res = 0;
    int now = -1;
    sort(s + 1, s + cnt + 1);
    for (int i = 1; i <= cnt; i ++)
    {
        int r = *(--s[i].end()), l = *s[i].begin();
        if(now == -1)
        {
            now = r;
        }
        else
        {
            if(l < now) res ++;
            now = max(r, now);
        }
        
    }
    printf("%d
", res);
}