Problem 1. Dance Mooves

Farmer John’s cows are showing off their new dance mooves!

At first, all NN cows (2≤N≤1052≤N≤105) stand in a line with cow ii in the iith position in line. The sequence of dance mooves is given by KK (1≤K≤2⋅1051≤K≤2⋅105) pairs of positions (a1,b1),(a2,b2),…,(aK,bK)(a1,b1),(a2,b2),…,(aK,bK). In each minute i=1…Ki=1…K of the dance, the cows in positions aiai and bibi in line swap. The same KK swaps happen again in minutes K+1…2KK+1…2K, again in minutes 2K+1…3K2K+1…3K, and so on, continuing indefinitely in a cyclic fashion. In other words,

In minute 11, the cows at positions a1a1 and b1b1 swap.
In minute 22, the cows at positions a2a2 and b2b2 swap.
...
In minute KK, the cows in positions aKaK and bKbK swap.
In minute K+1K+1, the cows in positions a1a1 and b1b1 swap.
In minute K+2K+2, the cows in positions a2a2 and b2b2 swap.
and so on ...

For each cow, please determine the number of unique positions in the line she will ever occupy.

INPUT FORMAT (input arrives from the terminal / stdin):

The first line contains integers NN and KK. Each of the next KK lines contains (a1,b1)…(aK,bK)(a1,b1)…(aK,bK) (1≤ai<bi≤N1≤ai<bi≤N).

OUTPUT FORMAT (print output to the terminal / stdout):

Print NN lines of output, where the iith line contains the number of unique positions that cow ii reaches.

SAMPLE INPUT:

SAMPLE OUTPUT:

Cow 11 reaches positions {1,2,3,4}{1,2,3,4}.
Cow 22 reaches positions {1,2,3,4}{1,2,3,4}.
Cow 33 reaches positions {1,2,3}{1,2,3}.
Cow 44 reaches positions {1,2,3,4}{1,2,3,4}.
Cow 55 never moves, so she never leaves position 55.

SCORING:

Test cases 1-5 satisfy N≤100,K≤200N≤100,K≤200.
Test cases 6-10 satisfy N≤2000,K≤4000N≤2000,K≤4000.
Test cases 11-20 satisfy no additional constraints.

解题思路

我们声称同一周期中所有奶牛的答案都是相同的。因为每进行 K 次交换，一切都会重复，如果两头奶牛在 K 次交换后一直处于同一位置，那么它们最终会到达相同的位置。 K 次交换后，牛 i 到位置 pi，所以牛 i 和牛 pi 的答案是一样的。这个逻辑扩展到循环中的每一头奶牛（奶牛 pi 的答案等于奶牛 pi 等等）。

#include <bits/stdc++.h>
using namespace std;int N,K;
int A[200001],B[200001]; //input
int P[100001]; //as described in analysis
vector<int>S[100001]; //as described in analysis
int from[100001]; //from[i] = where the cow in position i originated from
int cnt[100001]; //array to keep track of uniquePos
int uniquePos; //# of unique reachable positions//add in S_node
void add(int node){for (int x:S[node]){if (cnt[x]==0)uniquePos++;cnt[x]++;}
}//remove S_node
void remove(int node){for (int x:S[node]){if (cnt[x]==1)uniquePos--;cnt[x]--;}
}bool vis[100001];
queue<int>q; //stores all nodes in current cyclevoid dfs(int node){vis[node]=true;add(node);q.push(node);if (!vis[P[node]])dfs(P[node]);
}int main(){cin>>N>>K;for (int i=0;i<K;i++)cin>>A[i]>>B[i];//initialize from and Sfor (int i=1;i<=N;i++){from[i]=i;S[i].push_back(i);}//simulate the first K swaps, keeping track of where each position can reachfor (int i=0;i<K;i++){S[from[A[i]]].push_back(B[i]);S[from[B[i]]].push_back(A[i]);swap(from[A[i]],from[B[i]]);}//compute array P after first K swapsfor (int i=1;i<=N;i++)P[from[i]]=i;int ans[100001];//run a DFS on each cyclefor (int i=1;i<=N;i++)if (!vis[i]){dfs(i);int tempAns=uniquePos; //the answer //assign the answer for all nodes in the cycle, which we've stored in this queuewhile (!q.empty()){remove(q.front());ans[q.front()]=tempAns;q.pop();}}for (int i=1;i<=N;i++)cout<<ans[i]<<endl;return 0;
}

Problem 2. No Time to Paint

Bessie has recently received a painting set, and she wants to paint the long fence at one end of her pasture. The fence consists of NN consecutive 1-meter segments (1≤N≤1051≤N≤105). Bessie has 26 different colors available, which she labels with the letters 'A' through 'Z' in increasing order of darkness ('A' is a very light color, and 'Z' is very dark). She can therefore describe the desired color she wants to paint each fence segment as a length-NN string where each character is a letter.

Initially, all fence segments are uncolored. Bessie can color any contiguous range of segments with a single color in a single brush stroke as long as she never paints a lighter color over a darker color (she can only paint darker colors over lighter colors).

For example, an initially uncolored segment of length four can be colored as follows:

.... -> BBB. -> BBLL -> BQQL

Running short on time, Bessie thinks she may need to leave some consecutive range of fence segments unpainted! Currently, she is considering QQ candidate ranges (1≤Q≤1051≤Q≤105), each described by by two integers (a,b)(a,b) with 1≤a≤b≤N1≤a≤b≤N giving the indices of endpoints of the range a…ba…b of segments to be left unpainted.

For each candidate range, what is the minimum number of strokes needed to paint every fence segment outside those in the range with its desired color while leaving all fence segments inside the range uncolored? Note that Bessie does not actually do any painting during this process, so the answers for each candidate range are independent.

INPUT FORMAT (input arrives from the terminal / stdin):

The first line contains NN and QQ.

The next line contains a string of length NN characters representing the desired color for each fence segment.

The next QQ lines each contain two space-separated integers aa and bb representing a candidate range to possibly leave unpainted.

OUTPUT FORMAT (print output to the terminal / stdout):

For each of the QQ candidates, output the answer on a new line.

SAMPLE INPUT:

8 2
ABBAABCB
3 6
1 4

SAMPLE OUTPUT:

4
3

In this example, excluding the sub-range corresponding to the desired pattern

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USACO 1月 2020-2021 January Contest Silver 题解

Problem 1. Dance Mooves

解题思路

Problem 2. No Time to Paint

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