Good Bye 2017 E. New Year and Entity Enumeration

先按照绿点进行分块
第一个绿点和最后一个绿点之后很好处理不说了

两个绿点之间的讨论：
有两种方案
1：红(蓝)点和绿点顺序连接，距离为相邻绿点距离(也就是双倍绿点距离)
2：红(蓝)点和绿点的点阵中寻找最大的距离边，不连这一条，其他都顺序连，当然这样不连通，最后再绿点连接。(一个绿点距离+红(蓝)点阵处理 可以看到样例就是这样做的)

#include<iostream>

#include<map>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<set>

#include<vector>

#include<queue>

#include<stack>

#include<cmath>

#include<algorithm>

using namespace std;

typedef long long ll;

const int INF = 0x3f3f3f3f;

const int N = 3e5+5;

#define MS(x,y) memset(x,y,sizeof(x))

#define MP(x, y) make_pair(x, y)

int dis[N]; int color[N];

vector<int> so;

int main() {

    int n;

    while(~scanf("%d", &n)) {

        so.clear();

        for(int i = 0; i < n; ++i) {

            int a; char b[10];

            scanf("%d %s", &a, b);

            dis[i] = a;

            color[i] = b[0]=='G'? 3 : (b[0]=='R'? 1:2);

        }

        int ans = 0;

        for(int i = 0; i < n; ++i) {

            if(color[i] == 3) {

                so.push_back(i);

            }

        }

        if(so.empty()) {

            int st = -1, ed = 0;

            for(int i = 0; i < n; ++i) {

                if(color[i] == 1 && st == -1) st = i;

                if(color[i] == 1) ed = i;

            }

            if(st != -1) ans += dis[ed] - dis[st];

            st = -1; ed = 0;

            for(int i = 0; i < n; ++i) {

                if(color[i] == 2 && st == -1) st = i;

                if(color[i] == 2) ed = i;

            }

            if(st != -1) ans += dis[ed] - dis[st];

            printf("%d\n", ans);

            continue;

        }

        //1

        for(int i = 0; i < so[0]; ++i) {

            if(color[i] == 1)   {

                ans += dis[so[0]] - dis[i]; break;

            }

        }

        for(int i = 0; i < so[0]; ++i) {

            if(color[i] == 2)   {

                ans += dis[so[0]] - dis[i]; break;

            }

        }

        //2

        for(int i = 0; i <= so.size()-2; ++i) {

            int len = dis[so[i+1]] - dis[so[i]];

            int tt = 0;

            int maxx = 0; int pre = so[i];

            for(int j = so[i] + 1; j <= so[i+1]; ++j) {

                if(color[j] != 1 ) {

                //  printf("%d\n", j);

                    maxx = max(maxx, dis[j] - dis[pre]);

                    pre = j;

                }

            }

            tt += dis[so[i + 1]] - dis[so[i]] - maxx;

            maxx = 0; pre = so[i];

            for(int j = so[i] + 1; j <= so[i+1]; ++j) {

                if(color[j] != 2) {

                    maxx = max(maxx, dis[j] - dis[pre]);

                    pre = j;

                }

            }

            tt += dis[so[i + 1]] - dis[so[i]] - maxx;

            if(tt > len) tt = len;

            tt += len;

            ans += tt;

        }

        //3

        for(int i = n-1; i > so[so.size()-1]; --i) {

            if(color[i] == 1) {

                ans += dis[i] - dis[so[so.size()-1]]; break;

            }

        }

        for(int i = n-1; i > so[so.size()-1]; --i) {

            if(color[i] == 2) {

                ans += dis[i] - dis[so[so.size()-1]]; break;

            }

        }

        printf("%d\n", ans);

    }

    return 0;

}