hdu4117.GRE Words

http://acm.hdu.edu.cn/showproblem.php?pid=4117

题意，给一组固定顺序的字符串，每个字符串有一个价值，现在要你去除一些字符串，使得(1)剩下的字符串序列满足相邻的两个前一个是后一个的字串;(2)满足(1)的情况下剩下的字符串的价值和最大。

先说明一下：hdu的数据有可能含有非小写字母字符……

思路，把字符串做个自动机和按顺序合成一个大串，那么大串的每一点都对应自动机上的一个点，然后dp：dp[i][x]表示到大串的第i个字符，自动机上的第x的节点的最大价值，则dp[i][x] = max(dp[j1][x], dp[j2][fa[x]], dp[j3][fail[x]], ...) + val[i]就是说有点像在自动机上面dp，不过dp的顺序是按照大串的顺序。fa[x] 是自动机上x的入节点，fail[x]是x的失败指针指向的节点，要一值往前扫到0节点。val[i]是第i个点的价值，如果这个点某个小串的结尾，则这个点的价值就是小串的价值，否则是0。因为从i可以得出唯一的x，所以这个dp其实是一维，建议用x，因为转移时可以直接确定fa[x]等节点。理论时间复杂度为O(M^2), M是大串长度，但是如果加个判断如果fail[x] == fa[x], 不必往前扫，时间复杂都可以降到O(M^1.5).

交到hdu上面的结果很诡异：

// hdu4117. GRE Words - TRIE + dp#include <map>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define sqr(x) ((x) * (x))#define two(x) (1 << (x))#define X first#define Y secondtypedef long long LL;const int MAXN = 300010;const double eps = 1e-9;const int INF = 1000000000;struct NODE{map<char, int> s;int pre, fa;char c;} trie[MAXN];int n, m, tn, head, tail, ans, tch;int val[MAXN], node[MAXN], que[MAXN], f[MAXN];void build(int x, int fa, char c){trie[x].s.clear();trie[x].pre = -1;trie[x].fa = fa;trie[x].c = c;if (fa != -1)trie[fa].s[c] = x;}void insert(char str[]){int x = 0;for (int i = 0; str[i]; ++i){char c = str[i];if (trie[x].s.find(c) == trie[x].s.end())build(tn++, x, c);node[m + i] = x = trie[x].s[c];}}void bfs(){que[head = 0] = 0;tail = 1;while (head < tail){int x = que[head], c = trie[x].c;if (x == 0) trie[x].pre = -1;else{int y = trie[trie[x].fa].pre;while (y != -1 && trie[y].s.find(c) == trie[y].s.end()) y = trie[y].pre;trie[x].pre = (y == -1)? 0: trie[y].s[c];}for (map<char, int>::iterator i = trie[x].s.begin(); i != trie[x].s.end(); ++i){que[tail++] = i->second;}++head;}}void init(){char str[MAXN];m = 0;memset(val, 0, sizeof(val));build(0, -1, -1);tn = 1;scanf("%d", &n);for (int i = 0; i < n; ++i){int pts, l;scanf("%s%d", str, &pts); l = strlen(str);val[m + l - 1] = pts;insert(str);m += l;}bfs();}void work(){ans = 0;memset(f, 0, sizeof(f));for (int i = 0; i < m; ++i){int x = node[i];f[x] = max(f[x], max(f[x], f[trie[x].fa]) + val[i]);for (int p = trie[x].pre; p != -1 && trie[p].fa != trie[p].pre; p = trie[p].pre) //优化的地方f[x] = max(f[x], f[p] + val[i]);ans = max(ans, f[x]);}}int main(){int T, ca = 0;scanf("%d", &T);while (T--){init();work();printf("Case #%d: %d\n", ++ca, ans);}    return 0;}