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
; int color
;

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;
}