PAT A1039

本来以为的送分题变成了送命题

题目

一开始写的：

#include<iostream>
#include<vector>
#include <queue>
#include <algorithm>
#include<cstdio>
#include <map>
using namespace std;
//二分+前缀和
int num[100010];
int sum[100020];
map<string,vector<int> > mmp;
int main() {
    int class_num;
    int stu_num;
    scanf("%d %d",&stu_num,&class_num);
    for (int i = 0; i < class_num; ++i) {
        int now_class;
        int now_num;
        scanf("%d %d",&now_class,&now_num);
        for (int j = 0; j < now_num; ++j) {
            string temp;
            cin>>temp;
            mmp[temp].push_back(now_class);
        }


    }
    for (int i = 0; i < stu_num; ++i) {
        string s;
        cin>>s;
        cout<<s<<" ";
        int len=mmp[s].size();
        sort(mmp[s].begin(),mmp[s].begin()+len);
        printf("%d",len);
        for (int j = 0; j < len; ++j) {
            printf(" %d",mmp[s][j]);

        }
        cout<<endl;

    }



}

结果看了一眼书，书上说最后一组用map+string会超时，，，ok

其实你仔细看看哈，题目中其实有一个地方跟其他的不一样

这名字这么整齐划一，怎么看怎么像是个条件哦

所以，这个题，想要不超时

就从这个条件

不难想hash

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
int getid(char *name) {
    int id = 0;
    for(int i = 0; i < 3; i++)
        id = 26 * id + (name[i] - 'A');//其实你可以想象成26进制数
    id = id * 10 + (name[3] - '0');
    return id;
}
const int maxn = 26 * 26 * 26 * 10 + 10;
vector<int> v[maxn];

int main() {
    int n, k, no, num, id = 0;
    char name[5];
    scanf("%d %d", &n, &k);
    for(int i = 0; i < k; i++) {
        scanf("%d %d", &no, &num);
        for(int j = 0; j < num; j++) {
            scanf("%s", name);
            id = getid(name);
            v[id].push_back(no);
        }
    }
    for(int i = 0; i < n; i++) {
        scanf("%s", name);
        id = getid(name);
        sort(v[id].begin(), v[id].end());
        printf("%s %lu", name, v[id].size());
        for(int j = 0; j < v[id].size(); j++)
            printf(" %d", v[id][j]);
        printf("
");
    }
    return 0;
}