LightOJ 1044 - Palindrome Partitioning（dp）

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题目链接：LightOJ 1044 - Palindrome Partitioning

代码

#include <cstdio>#include <cstring>#include <queue>#include <vector>#include <algorithm>using namespace std;const int maxn = 1005;const int inf = 0x3f3f3f3f;typedef pair<int,int> pii;char S[maxn];int N, dp[maxn];vector<int> G[maxn];void init () {    scanf("%s", S + 1);    N = strlen(S + 1);    for (int i = 0; i <= N; i++)        G[i].clear();    queue<pii> que;    for (int i = 1; i <= N; i++)        que.push(make_pair(i, i));    for (int i = 1; i < N; i++) if (S[i] == S[i+1])        que.push(make_pair(i, i + 1));    while (!que.empty()) {        int l = que.front().first;        int r = que.front().second;        que.pop();        G[l-1].push_back(r);        l--, r++;        if (l == 0 || r > N) continue;        if (S[l] == S[r])            que.push(make_pair(l, r));    }}int solve () {    memset(dp, inf, sizeof(dp));    dp[0] = 0;    for (int i = 0; i < N; i++) {        for (int j = 0; j < G[i].size(); j++)            dp[G[i][j]] = min(dp[G[i][j]], dp[i] + 1);    }    return dp[N];}int main () {    int cas;    scanf("%d", &cas);    for (int kcas = 1; kcas <= cas; kcas++) {        init();        printf("Case %d: %d\n", kcas, solve());    }    return 0;}

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