Ural 1542. Autocompletion（二分）

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题目链接：点击打开链接

思路：

因为单词最长15，我们把每个单词不同长度的前缀存起来，排序之后二分即可，复杂度O(nlogn)

细节参见代码：

#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;// & 0x7FFFFFFFconst 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;}

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