E. Byteland, Berland and Disputed Cities
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
The cities of Byteland and Berland are located on the axis Ox. In addition, on this axis there are also disputed cities, which belong to each of the countries in their opinion. Thus, on the line Ox there are three types of cities:
the cities of Byteland,
the cities of Berland,
disputed cities.
Recently, the project BNET has been launched — a computer network of a new generation. Now the task of the both countries is to connect the cities so that the network of this country is connected.
The countries agreed to connect the pairs of cities with BNET cables in such a way that:
If you look at the only cities of Byteland and the disputed cities, then in the resulting set of cities, any city should be reachable from any other one by one or more cables,
If you look at the only cities of Berland and the disputed cities, then in the resulting set of cities, any city should be reachable from any other one by one or more cables.
Thus, it is necessary to choose a set of pairs of cities to connect by cables in such a way that both conditions are satisfied simultaneously. Cables allow bi-directional data transfer. Each cable connects exactly two distinct cities.
The cost of laying a cable from one city to another is equal to the distance between them. Find the minimum total cost of laying a set of cables so that two subsets of cities (Byteland and disputed cities, Berland and disputed cities) are connected.
Each city is a point on the line Ox. It is technically possible to connect the cities a and b with a cable so that the city c (a<c<b) is not connected to this cable, where a, b and c are simultaneously coordinates of the cities a, b and c.
Input
The first line contains a single integer n (2≤n≤2⋅105) — the number of cities.
The following n lines contains an integer xi and the letter ci (−109≤xi≤109) — the coordinate of the city and its type. If the city belongs to Byteland, ci equals to 'B'. If the city belongs to Berland, ci equals to «R». If the city is disputed, ci equals to 'P'.
All cities have distinct coordinates. Guaranteed, that the cities are given in the increasing order of their coordinates.
Output
Print the minimal total length of such set of cables, that if we delete all Berland cities (ci='R'), it will be possible to find a way from any remaining city to any other remaining city, moving only by cables. Similarly, if we delete all Byteland cities (ci='B'), it will be possible to find a way from any remaining city to any other remaining city, moving only by cables.
Examples
inputCopy
4
-5 R
0 P
3 P
7 B
outputCopy
12
inputCopy
5
10 R
14 B
16 B
21 R
32 R
outputCopy
24
Note
In the first example, you should connect the first city with the second, the second with the third, and the third with the fourth. The total length of the cables will be 5+3+4=12.
In the second example there are no disputed cities, so you need to connect all the neighboring cities of Byteland and all the neighboring cities of Berland. The cities of Berland have coordinates 10,21,32, so to connect them you need two cables of length 11 and 11. The cities of Byteland have coordinates 14 and 16, so to connect them you need one cable of length 2. Thus, the total length of all cables is 11+11+2=24.
题意：
在坐标轴上有3种城市，分别是R，B，P，让你给一些城市连上无向边 满足这2个条件：
1、 删除所有的R节点之后，剩下的节点两两相互连通。
2、 删除所有的B节点之后，剩下的节点两两相互连通。
如果连接边的成本是两个城市的距离，请你是算出最小的距离。
思路： 对于每一个B节点，我们肯定是要让其与上一个B节点连接，R也相同， 当遇到P 节点使，有两种选择：1、 让其与上一个B和上一个R相连。
2、让其与上一个P相连，然后删除一条最长的B-B或者R-R边。 这样一定可以满足题目要求的条件。
我们当然要选择成本最小的一个。
注意每一次选择第2个的时候，把维护的最长的B-B或者R-R边清空。
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d
",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2) { ans = ans * a % MOD; } a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int *p);
const int maxn = 1000010;
const ll inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll ans = 0ll;
ll preb = -inf;
ll prer = -inf;
ll prep = -inf;
ll max_r = 0ll;
ll max_b = 0ll;
int n;
ll pos;
char t;
int main()
{
//freopen("D:\code\text\input.txt","r",stdin);
//freopen("D:\code\text\output.txt","w",stdout);
gbtb;
cin >> n;
repd(i, 1, n) {
cin >> pos >> t;
if (t == 'R' || t == 'P') {
if (prer != -inf) {
ans += pos - prer;
max_r = max(max_r, pos - prer);
}
prer = pos;
}
if (t == 'B' || t == 'P') {
if (preb != -inf) {
ans += pos - preb;
max_b = max(max_b, pos - preb);
}
preb = pos;
}
if (t == 'P') {
if(prep!=-inf)
{
ans+=min(0ll,-max_r-max_b+pos-prep);
}
prep=pos;
max_r=0;
max_b=0;
}
}
cout<<ans<<endl;
return 0;
}
inline void getInt(int *p)
{
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '
');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
} else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}