URAL 1181 Cutting a Painted Polygon

题目

对于一个给定长度的字符串，这个字符串满足：

1.有且仅有RGB三个字母，且每个字母至少出现一次

2.相邻的两个字母不同（环形）

题解

如果只有两种颜色，那么显然无解

否则必定可以分一下两种情况：

1、有一种颜色只有1个，那么从这个连边到所有其他点分割

2、否则可以找到连续的三个点，它们的颜色互不相同

然后把这三个点连成三角形，变成n-1的问题，重复步骤

code

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>

const int maxn = 1010;

typedef std::vector<int> Polygon;

int n;
char s[maxn];

int nxt(int i, int n) { return (i+1)%n; }

void dfs(const Polygon& A) {
//  printf("L, R: %d %d\n", L, R);
int n = A.size();
//  for (int i = 0; i < n; ++ i) printf("%d ", A[i]); putchar('\n');
int r = 0, g = 0, b = 0, pr, pg, pb, p3;

for (int i = 0; i < n; ++ i) {
if (s[A[i]] == 'R') { ++ r; pr = i; }
if (s[A[i]] == 'G') { ++ g; pg = i; }
if (s[A[i]] == 'B') { ++ b; pb = i; }
if (s[A[i]] != s[A[nxt(i, n)]] && s[A[i]] != s[A[nxt(nxt(i, n), n)]] && s[A[nxt(i, n)]] != s[A[nxt(nxt(i, n), n)]]) { p3 = i; }
}

if (r == 1 || g == 1 || b == 1) {
int p;
if (r == 1) p = pr;
if (g == 1) p = pg;
if (b == 1) p = pb;
for (int i = nxt(nxt(p, n), n); nxt(i, n) != p; i = nxt(i, n)) printf("%d %d\n", A[p]+1, A[i]+1);
return;
}

printf("%d %d\n", A[p3]+1, A[nxt(nxt(p3, n), n)]+1);
Polygon B; for (int i = nxt(nxt(p3, n), n); i != nxt(p3, n); i = nxt(i, n)) B.push_back(A[i]);
dfs(B);
}

void solve() {
scanf("%d%s", &n, s);

/*  int r = 0, g = 0, b = 0;
for (int i = 0; i < n; ++ i) if (s[i] == s[(i+1)%n]) { puts("0"); return; }
for (int i = 0; i < n; ++ i) {
if (s[i] == 'R') { ++ r; }
if (s[i] == 'G') { ++ g; }
if (s[i] == 'B') { ++ b; }
}
if (r == 0 || g == 0 || b == 0) { puts("0"); return; }
*/
printf("%d\n", n-3);

Polygon S; for (int i = 0; i < n; ++ i) S.push_back(i);
dfs(S);
}

int main() {
//  freopen("G.in", "r", stdin);

solve();
//  for(;;);
return 0;
}