Codeforces Round #378 (Div. 2) C. Epidemic in Monstropolis(模拟)

题意：给你一串长度为n数列，数字大的可以吃掉比他小的数字（要严格大于）然后这两个数合并成一个数，更新队伍。再给你一串长度为k的数列。

让你进行上述操作使原来序列变成现在这串数。如果不能输出“NO”，可以的话输出“YES”然后输出操作过程，比如x位上

的数吃掉左边的则输出x L，右边的则是x R，不吃则不输出。当然结果可能有多个，输出任意一个。

每个bi对应一个a数组的区间，要成立的条件是每个区间要有一个最大的能吃光周围的。一个最大的一开始可以往右或往左（但不能和他一样大），确定好一开始向哪个方向后，他

可以往这个方向吃到底，因为他是最大的，然后再反方向把其他的吃完即可。重要的是怎么记录路径，往左吃下标是每次-1，往右吃是下标不变。

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 505;
int b[maxn], n, k, cur;
struct node
{
int id, cmd;
node() {}
node(int ii, int cc): id(ii), cmd(cc) {}
}ans[maxn];

bool solve(int *a, int len, int pre)
{
//如果只有一个数肯定是可以的
if(len == 1) return 1;
int i, maxEle = -1;
for(i = 1; i < len; i++)
if(a[i] != a[i-1]) break;
//如果这个区间所有值都相同，肯定不可以合并
if(i >= len) return 0;
for(int i = 0; i < len; i++)
maxEle = max(maxEle, a[i]);
int s = -1;
for(int i = 0; i < len; i++)
{
if(a[i] == maxEle)
{
//如果最大值在开头或结尾注意特判下
if(i == 0 && a[1] != maxEle) { s = 0; break; }
else if(i == len-1 && a[len-2] != maxEle) { s = len-1; break; }
else if(i && a[i-1] != maxEle || a[i+1] != maxEle) { s = i; break; }
else;
}
}

//注意要判断下这个起点是先往右还是先往左
if(s > 0 && a[s] == a[s-1])
{
for(int i = 0; i < len-s-1; i++)
ans[cur++] = node(pre+s+1, 1);
for(int i = 0; i < s; i++)
ans[cur++] = node(pre+s+1-i, 0);
}
else
{
for(int i = 0; i < s; i++)
ans[cur++] = node(pre+s+1-i, 0);
for(int i = 0; i < len-s-1; i++)
ans[cur++] = node(pre+1, 1);
}

return 1;
}
int main(void)
{
int a[maxn];
while(cin >> n)
{
cur = 0;
for(int i = 0; i < n; i++)
scanf("%d", &a[i]);
scanf("%d", &k);
for(int i = 0; i < k; i++)
scanf("%d", &b[i]);
int t = 0, finish = 0, num = 0, num2 = 0;
for(int i = 0; i < n; i++)
{
num++;
t += a[i];
if(t == b[num2])
{
if(!solve(a+i-num+1, num, num2)) break;
finish += num;
t = num = 0;
num2++;
}
}
//要保证用完a、b数组
if(finish != n || num2 != k) puts("NO");
else
{
puts("YES");
for(int i = 0; i < cur; i++)
printf("%d %c\n", ans[i].id, ans[i].cmd ? 'R' : 'L');
}

}
return 0;
}


C. Epidemic in Monstropolis

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

There was an epidemic in Monstropolis and all monsters became sick. To recover, all monsters lined up in queue for an appointment to the only doctor in the city.

Soon, monsters became hungry and began to eat each other.

One monster can eat other monster if its weight is strictly greater than the weight of the monster being eaten, and they stand in the queue next to each other. Monsters eat each other instantly.
There are no monsters which are being eaten at the same moment. After the monster A eats the monster B,
the weight of the monster A increases by the weight of the eaten monster B.
In result of such eating the length of the queue decreases by one, all monsters after the eaten one step forward so that there is no empty places in the queue again. A monster can eat several monsters one after another. Initially there were n monsters
in the queue, the i-th of which had weight ai.

For example, if weights are [1, 2, 2, 2, 1, 2] (in order of queue, monsters are numbered from 1 to 6 from
left to right) then some of the options are:

the first monster can't eat the second monster because a1 = 1 is
not greater than a2 = 2;

the second monster can't eat the third monster because a2 = 2 is
not greater than a3 = 2;

the second monster can't eat the fifth monster because they are not neighbors;

the second monster can eat the first monster, the queue will be transformed to [3, 2, 2, 1, 2].

After some time, someone said a good joke and all monsters recovered. At that moment there were k (k ≤ n)
monsters in the queue, thej-th of which had weight bj.
Both sequences (a and b)
contain the weights of the monsters in the order from the first to the last.

You are required to provide one of the possible orders of eating monsters which led to the current queue, or to determine that this could not happen. Assume that the doctor didn't make any appointments while monsters were eating each other.

Input

The first line contains single integer n (1 ≤ n ≤ 500) —
the number of monsters in the initial queue.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106) —
the initial weights of the monsters.

The third line contains single integer k (1 ≤ k ≤ n) —
the number of monsters in the queue after the joke.

The fourth line contains k integers b1, b2, ..., bk (1 ≤ bj ≤ 5·108) —
the weights of the monsters after the joke.

Monsters are listed in the order from the beginning of the queue to the end.

Output

In case if no actions could lead to the final queue, print "NO" (without quotes) in the only line.

Otherwise print "YES" (without quotes) in the first line. In the next n - k lines print actions in the chronological order.
In each line printx — the index number of the monster in the current queue which eats and, separated by space, the symbol 'L'
if the monster which stays the x-th in the queue eats the monster in front of him, or 'R'
if the monster which stays the x-th in the queue eats the monster behind him. After each eating the queue is enumerated again.

When one monster eats another the queue decreases. If there are several answers, print any of them.

Examples

input

6
1 2 2 2 1 2
2
5 5

output

YES
2 L
1 R
4 L
3 L

input

5
1 2 3 4 5
1
15

output

YES
5 L
4 L
3 L
2 L

input

5
1 1 1 3 3
3
2 1 6

output

NO

Note

In the first example, initially there were n = 6 monsters, their weights are [1, 2, 2, 2, 1, 2] (in
order of queue from the first monster to the last monster). The final queue should be [5, 5]. The following sequence of eatings leads to the
final queue:

the second monster eats the monster to the left (i.e. the first monster), queue becomes [3, 2, 2, 1, 2];

the first monster (note, it was the second on the previous step) eats the monster to the right (i.e. the second monster), queue becomes [5, 2, 1, 2];

the fourth monster eats the mosnter to the left (i.e. the third monster), queue becomes [5, 2, 3];

the finally, the third monster eats the monster to the left (i.e. the second monster), queue becomes [5, 5].

Note that for each step the output contains numbers of the monsters in their current order in the queue.