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Gym 100962J Jimi Hendrix (树形DP)
题意:给定一棵树,然后每条边有一个字母,然后给定一行字符串,问你能不能从这棵树上找到,并输出两个端点。
析:树形DP,先进行递归到叶子结点,然后再回溯,在回溯的时候要四个值,一个是正着匹配的长度和端点,一个是反着匹配的长度和端点,
然后一个一个匹配,并不断更新这个长度和端点。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>#define print(a) printf("%d\n", (a))#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 5e5 + 5;const int mod = 1e9 + 7;const int dr[] = {-1, 0, 1, 0};const int dc[] = {0, 1, 0, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m;}vector<char> w[maxn];vector<int> G[maxn];int ansx, ansy;struct Node{ int lans, rans; int l, r;};Node dp[maxn];char s[maxn];bool dfs(int u, int fa){ for(int i = 0; i < G[u].size(); ++i){ int v = G[u][i]; if(v == fa) continue; if(dfs(v, u)) return true; int l = s[dp[v].l+1] == w[u][i] ? dp[v].l+1 : dp[v].l; int r = s[m-dp[v].r] == w[u][i] ? dp[v].r+1 : dp[v].r; if(l + dp[u].r >= m){ ansx = dp[v].lans; ansy = dp[u].rans; return true; } else if(r + dp[u].l >= m){ ansx = dp[u].lans; ansy = dp[v].rans; return true; } if(l > dp[u].l){ dp[u].l = l; dp[u].lans = dp[v].lans; } if(r > dp[u].r){ dp[u].r = r; dp[u].rans = dp[v].rans; } } return false;}int main(){ while(scanf("%d %d", &n, &m) == 2){ for(int i = 1; i <= n; ++i) G[i].clear(), w[i].clear(); int u, v; char ch; for(int i = 1; i < n; ++i){ scanf("%d %d %c", &u, &v, &ch); dp[i].l = dp[i].r = 0; dp[i].lans = dp[i].rans = i; w[u].push_back(ch); w[v].push_back(ch); G[u].push_back(v); G[v].push_back(u); } dp[n].l = dp[n].r = 0; dp[n].lans = dp[n].rans = n; scanf("%s", s+1); ansx = ansy = -1; dfs(1, -1); printf("%d %d\n", ansx, ansy); } return 0;}
Gym 100962J Jimi Hendrix (树形DP)
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