HDU 1298 - T9(trie树)

来源：互联网发布：网络编程实用教程第3版编辑：程序博客网时间：2024/06/04 19:01

题目链接 HDU1298

【题意】给出n(<=1000)个单词以及其出现概率值，每个单词的出现概率又是所有到这个位置的前缀之和，比如

or 2,np 1,nh 3，则串n出现概率为1+3,nh为3，np为1，o为2，or为2；然后求出每次输入手机键盘按钮时对应的概率最大的子串(不一定要出现的字符串，他们前缀都可以)。

【分析】一开始题意理解错了，以为概率为路径之和，知道题意后，还是蛮好写的，只要把字典树的查询改一下，用dfs完成，每次控制下可以查询的字符串(手机键盘数字对应的3，4个字母)就可以暴搜了。

#include <cstdio>#include <cstring>#include <cctype>#include <cmath>#include <map>//#include <unordered_map>#include <queue>#include <stack>#include <vector>#include <string>#include <algorithm>using namespace std;typedef long long LL;#define rep(i,a,n) for(int i = a; i < n; i++)#define repe(i,a,n) for(int i = a; i <= n; i++)#define per(i,n,a) for(int i = n; i >= a; i--)#define clc(a,b) memset(a,b,sizeof(a))const int INF = 0x3f3f3f3f, MAXN = 1010;int tree[MAXN<<5][30], val[MAXN<<5], cnt;void init(){clc(tree[0],-1);cnt = 1;}void insert(const char *s, int v){int u = 0, n = strlen(s);rep(i,0,n){int c = s[i]-'a';if(-1 == tree[u][c]){clc(tree[cnt],-1);val[cnt] = 0;tree[u][c] = cnt++;}u = tree[u][c];val[u] += v;}}void get_num(int &si, int &len, int num){if(num <= 6){si = (num-2)*3;len = 3;return;}if(7 == num){si = 15;len = 4;return;}if(8 == num){si = 19;len = 3;return;}si = 22, len = 4;}char pt[110], ans[110];int mx;void query(char *s, int u, int cnt, int len){if(cnt == len){if(mx < val[u]){strcpy(ans,pt);mx = val[u];}return;}//得到按键对应的字母int si, n;get_num(si,n,s[cnt]-'0');rep(i,0,n){int c = si+i;pt[cnt] = c+'a', pt[cnt+1] = 0;if(-1 == tree[u][c]) continue;query(s,tree[u][c], cnt+1, len);}return;}int main(){#ifdef SHYfreopen("e:\\1.txt", "r", stdin);//freopen("e:\\out.txt","w",stdout);#endifint t, count = 0;scanf("%d%*c", &t);while(t--){int n, v;char s[110];scanf("%d%*c", &n);init();rep(i,0,n){scanf("%s %d%*c",s, &v);insert(s,v);}int m;scanf("%d%*c", &m);printf("Scenario #%d:\n", ++count);rep(i,0,m){scanf("%s", s);int len = strlen(s);s[len-1] = 0;bool end = false;rep(i,1,len){if(!end){mx = 0;query(s,0,0,i);if(mx)puts(ans);elseputs("MANUALLY"), end = true;}elseputs("MANUALLY");}putchar('\n');}putchar('\n');}return 0;}

0 0