本文共 3682 字，大约阅读时间需要 12 分钟。
Roy and Biv have a set of n points on the infinite number line.
Each point has one of 3 colors: red, green, or blue.
Roy and Biv would like to connect all the points with some edges. Edges can be drawn between any of the two of the given points. The cost of an edge is equal to the distance between the two points it connects.
They want to do this in such a way that they will both see that all the points are connected (either directly or indirectly).
However, there is a catch: Roy cannot see the color red and Biv cannot see the color blue.
Therefore, they have to choose the edges in such a way that if all the red points are removed, the remaining blue and green points are connected (and similarly, if all the blue points are removed, the remaining red and green points are connected).
Help them compute the minimum cost way to choose edges to satisfy the above constraints.
The first line will contain an integer n (1 ≤ n ≤ 300 000), the number of points.
The next n lines will contain two tokens pi and ci (pi is an integer, 1 ≤ pi ≤ 109, ci is a uppercase English letter 'R', 'G' or 'B'), denoting the position of the i-th point and the color of the i-th point. 'R' means red, 'G' denotes green, and 'B' means blue. The positions will be in strictly increasing order.
Print a single integer, the minimum cost way to solve the problem.
41 G5 R10 B15 G
23
41 G2 R3 B10 G
12
发现错误没时间改了，就差几秒钟交不上去了。。
题意：一条横轴上有n个点，每个点都有一个颜色（RGB三种颜色中的一种），你可以将任意两点连接起来，代价为两点之间的距离，问如何以最小的代价满足：①删除所有红点(R)后剩下的点全部连通；②删除所有蓝色(B)点后剩下的点全部连通
思路：
首先很显然将一个红点和一个蓝点连起来没有意义
对于没有绿点的情况：将所有红点用一条直线连起来，将所有蓝点用一条直线连起来就ok
对于有绿点的情况，对于相邻的两个绿点有且只有两种连法：
两种取最优的，然后最两边的绿点外的所有红点蓝点全部连一条线到最近的绿点即可
#include#include using namespace std;#define LL long longtypedef struct{ int x; int tp;}Point;Point s[300005];int Min[4], Max[4];LL a1[300005];int main(void){ LL ans; char ch; int i, n, bet, last1, last2, cnt; scanf("%d", &n); for(i=1;i<=3;i++) Min[i] = 1e9+1, Max[i] = 0; for(i=1;i<=n;i++) { scanf("%d %c", &s[i].x, &ch); if(ch=='R') s[i].tp = 1, Max[1] = max(s[i].x, Max[1]), Min[1] = min(s[i].x, Min[1]); if(ch=='G') s[i].tp = 3, Max[3] = max(s[i].x, Max[3]), Min[3] = min(s[i].x, Min[3]); if(ch=='B') s[i].tp = 2, Max[2] = max(s[i].x, Max[2]), Min[2] = min(s[i].x, Min[2]); } if(Max[3]==0) { ans = 0; if(Max[1]) ans += Max[1]-Min[1]; if(Max[2]) ans += Max[2]-Min[2]; printf("%lld\n", ans); } else { ans = Max[3]-Min[3]; if(Max[1]) { cnt = bet = 0, last1 = last2 = -1; for(i=1;i<=n;i++) { if(s[i].tp==3) { if(last2!=-1) { ans += s[i].x-s[last2].x; bet = max(bet, s[i].x-s[last2].x); } if(last1!=-1) { if(bet) a1[++cnt] = s[i].x-s[last1].x-bet; ans -= bet; } last1 = last2 = i; bet = 0; } else if(s[i].tp==1) { if(last2!=-1) { ans += s[i].x-s[last2].x; bet = max(bet, s[i].x-s[last2].x); } last2 = i; } } } if(Max[3]) { cnt = bet = 0, last1 = last2 = -1; for(i=1;i<=n;i++) { if(s[i].tp==3) { if(last2!=-1) { ans += s[i].x-s[last2].x; bet = max(bet, s[i].x-s[last2].x); } if(last1!=-1) { if(bet) a1[++cnt] += s[i].x-s[last1].x-bet; if(a1[cnt]>s[i].x-s[last1].x) ans -= a1[cnt]-s[i].x+s[last1].x; ans -= bet; } last1 = last2 = i; bet = 0; } else if(s[i].tp==2) { if(last2!=-1) { ans += s[i].x-s[last2].x; bet = max(bet, s[i].x-s[last2].x); } last2 = i; } } } printf("%lld\n", ans); } return 0;}