hdu 5340 Three Palindromes（manachar）

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题目链接：hdu 5340 Three Palindromes

用manachar处理一下，分别找出所有包括起始位置和终止位置的回文串，然后枚举第一个串和第三个串，判断中间剩下的是否是回文。

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;typedef pair<int,int> pii;const int maxn = 50005;/********* mannachar *************/int rad[maxn];char str[maxn];int manachar(char* tmpstr) {int len = strlen(tmpstr), mv = 0;str[mv++] = '$';for (int i = 0; i <= len; i++) {str[mv++] = '#';str[mv++] = tmpstr[i];}int ans = 0, mix = 0, idx = 0;for (int i = 1; i <= mv; i++) {if (mix > i)rad[i] = min(rad[2 * idx - i], mix - i);elserad[i] = 1;for ( ; str[i - rad[i]] == str[i + rad[i]]; rad[i]++) {if (mix < i + rad[i]) {mix = i + rad[i];idx = i;}}ans = max(ans, rad[i]);}return ans - 1;}/*********************************/int N;vector<int> L, R;vector<pii> G;char s[maxn];bool solve () {N = strlen(s);//G.clear();L.clear();R.clear();for (int i = 0; i < N; i++) {int k = rad[i * 2 + 2] / 2 - 1;int l = i - k;int r = i + k;//G.push_back(make_pair(l + 1, r + 1));if (l == 0)L.push_back(r + 1);if (r == N - 1)R.push_back(l + 1);}for (int i = 1; i < N; i++) {int k = rad[i * 2 + 1] / 2;int l = i - k;int r = i + k - 1;if (l == 0)L.push_back(r + 1);if (r == N - 1)R.push_back(l + 1);}sort(L.begin(), L.end());sort(R.begin(), R.end());for (int i = 0; i < L.size(); i++) {for (int j = R.size() - 1; j >= 0; j--) {if (L[i] + 1 >= R[j])break;int k = L[i] + R[j];int l = rad[k] - 1;if (l >= R[j] - L[i] - 1) {//printf("%d %d %d\n", L[i], R[j], l);return true;}}}return false;}int main () {int cas;scanf("%d", &cas);while (cas--) {scanf("%s", s);manachar(s);printf("%s\n", solve() ? "Yes" : "No");}return 0;}

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