Ural 1542. Autocompletion（二分）

題目鏈接：點擊打開鏈接

思路：

因爲單詞最長15，  我們把每個單詞不同長度的前綴存起來， 排序之後二分即可，  複雜度O(nlogn)

細節參見代碼：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <ctime>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const double eps = 1e-6;
const double PI = acos(-1);
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
// & 0x7FFFFFFF
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 1e5 + 10;
int T,n,m,L[maxn];
struct node {
    char s[16], name[16];
    int p;
    node(char _s[], char str[], int _p) {
        strcpy(s, _s);
        strcpy(name, str);
        p = _p;
    }
    node() {}
    bool operator < (const node& rhs) const {
        int cur = strcmp(name, rhs.name);
        if(cur == 0) {
            if(p == rhs.p) {
                int cmp = strcmp(s, rhs.s);
                return cmp < 0;
            }
            else return p > rhs.p;
        }
        else return cur < 0;
    }
};
node a[maxn];
char op[19];
vector<node> b[17];
int main() {
    while(~scanf("%d", &n)) {
        int maxlen = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%s%d", a[i].s, &a[i].p);
            L[i] = strlen(a[i].s);
            maxlen = max(maxlen, L[i]);
        }
        for(int i = 1; i <= maxlen; i++) b[i].clear();
        for(int i = 1; i <= n; i++) {
            for(int j = 0; j < L[i]; j++) {
                op[j] = a[i].s[j];
                op[j+1] = 0;
                b[j+1].push_back(node(a[i].s, op, a[i].p));
            }
        }
        for(int i = 1; i <= maxlen; i++) sort(b[i].begin(), b[i].end());
        scanf("%d", &m);
        while(m--) {
            scanf("%s", op);
            int len = strlen(op);
            int pos = lower_bound(b[len].begin(), b[len].end(), node(op, op, INF)) - b[len].begin();
            int lenth = b[len].size();
            for(int i = pos; i < lenth; i++) {
                if(strcmp(b[len][i].name, op) != 0) break;
                printf("%s\n", b[len][i].s);
                if(i - pos + 1 >= 10) break;
            }
            if(m) printf("\n");
        }
    }
    return 0;
}