暴力+树状数组维护 Codeforces Round #378 (Div. 2) C

题目大意：给你一个长度为n的数组a，然后数值大的可以合并数值小的，且合并了以后该数组的长度-1.给你一个长度为k目标数组b，问，是否可以从a数组变到b数组，是就yes并且输出步骤。否就输出no

思路：因为合并的时候要删除某一个数字，很容易想到用树状数组来维护。然后分析一下就可以得到b中第i个数，一定是由a的某一段区间得到的。我们只要找到这一段区间就行啦。

 #include <bits/stdc++.h>
 using namespace std;
 #pragma comment(linker,"/STACK:102400000,102400000")
 #define LL long long
 #define ALL(a) a.begin(), a.end()
 #define pb push_back
 #define mk make_pair
 #define fi first
 #define se second
 #define haha printf("haha\n")
 const int maxn =  + ;
 int n, k;
 int a[maxn], b[maxn];
 int tree[maxn];
 vector<pair<int, int> > v;

 void update(int x, int val = ){
     for (int i = x; i <= maxn - ; i += i & -i)
         tree[i] += val;
 }

 int sum(int pos){
     int ans = ;
     for (int i = pos - ; i > ; i -= i & -i){
         ans += tree[i];
     }
     return ans;
 }

 vector<pair<int, char> > ans;

 void eat(int pos, int lb, int rb){
     if (pos == rb || a[pos] > a[pos - ]){
         for (int i = pos - ; i >= lb; i--){
             int reduce = sum(pos);
             ans.push_back(mk(pos - reduce, 'L'));
             update(i);
         }
         for (int i = pos + ; i <= rb; i++){
             int reduce = sum(pos);
             ans.push_back(mk(pos - reduce, 'R'));
             update(i);
         }
     }
     else {
         for (int i = pos + ; i <= rb; i++){
             int reduce = sum(pos);
             ans.push_back(mk(pos - reduce, 'R'));
             update(i);
         }
         for (int i = pos - ; i >= lb; i--){
             int reduce = sum(pos);
             ans.push_back(mk(pos - reduce, 'L'));
             update(i);
         }
     }
 }

 bool solve(){
     a[n + ] = 0x3f3f3f3f;
     int lb = ;
     for (int i = ; i <= k; i++){
         int val = b[i];
         for (int j = lb; j <= n; j++){
             if (val == a[j]){
                 v.push_back(mk(lb, j));
                 lb = j + ;
                 break;
             }
             else if (val > a[j]) val -= a[j];
             else return false;
         }
         if (lb > n) break;
     }
     if (v.size() != k) return false;
     for (int i = ; i < v.size(); i++){
         int lb = v[i].fi, rb = v[i].se;
         int val = a[lb];
         bool flag = (rb == lb ? true : false);
         for (int j = lb + ; j <= rb; j++){
             if (val != a[j]) {
                 flag = true;
                 break;
             }
         }
         if (!flag) return false;
     }
     for (int i = ; i < v.size(); i++){
         int lb = v[i].fi, rb = v[i].se;
         if (lb == rb) continue;
         int maxval = a[lb], pos = lb;
         for (int j = lb + ; j <= rb; j++){
             if (maxval < a[j]) {maxval = a[j]; pos = j;}
             else if (maxval == a[j] && a[j - ] < a[j]) pos = j;
             else if (maxval == a[j] && a[j + ] < a[j] && j < rb) pos = j;
         }
         eat(pos, lb, rb);
     }
     return true;
 }

 int main(){
     scanf("%d", &n);
     int sum = ;
     for (int i = ; i <= n; i++) scanf("%d", a + i), sum += a[i];
     scanf("%d", &k);
     for (int i = ; i <= k; i++) scanf("%d", b + i), sum -= b[i];
     if (sum != ) printf("NO\n");
     else {
         bool flag = solve();
         if (!flag) printf("NO\n");
         else {
             printf("YES\n");
             for (int i = ; i < ans.size(); i++){
                 pair<int, char> p = ans[i];
                 printf("%d %c\n", p.fi, p.se);
             }
         }
     }
     return ;
 }