hash思维题目

Problem 2280 Magic

Accept: 50    Submit: 156
Time Limit: 2000 mSec    Memory Limit : 262144 KB

 Problem Description

Kim is a magician, he can use n kinds of magic, number from 1 to n. We use string Si to describe magic i. Magic Si will make Wi points of damage. Note that Wi may change over time.

Kim obey the following rules to use magic:

Each turn, he picks out one magic, suppose that is magic Sk, then Kim will use all the magic i satisfying the following condition:

1. Wi<=Wk

2. Sk is a suffix of Si.

Now Kim wondering how many magic will he use each turn.

Note that all the strings are considered as a suffix of itself.

 Input

First line the number of test case T. (T<=6)

For each case, first line an integer n (1<=n<=1000) stand for the number of magic.

Next n lines, each line a string Si (Length of Si<=1000) and an integer Wi (1<=Wi<=1000), stand for magic i and it’s damage Wi.

Next line an integer Q (1<=Q<=80000), stand for there are Q operations. There are two kinds of operation.

“1 x y” means Wx is changed to y.

“2 x” means Kim has picked out magic x, and you should tell him how many magic he will use in this turn.

Note that different Si can be the same.

 Output

For each query, output the answer.

 Sample Input

15abracadabra 2adbra 1bra 3abr 3br 252 32 51 2 52 32 2

 Sample Output

3121

 Source

第八届福建省大学生程序设计竞赛-重现赛（感谢承办方厦门理工学院）

#include<algorithm>
#include<stdio.h>
#include<string>
#include<vector>
#include<iostream>
using namespace std;
const int maxn=10010;
vector<int>G[maxn];
struct qq
{

    string a;
    int b;

} q[maxn];
int main()
{
    string w1="adbra",w2="dabra";
    if(w1==w2)
        printf("yes\n");

//    string ww;
//    cin>>ww;
//    cout<<ww.substr(2,2);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        for(int i=0; i<1000; i++)
        {
            G[i].clear();
            q[i].b=0;
            q[i].a="";
        }
        int n;
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            cin>>q[i].a>>q[i].b;

        }
        for(int i=1; i<n; i++)
        {

            for(int j=i+1; j<=n; j++)
            {
                if(q[i].a.size()>q[j].a.size())
                {
                    int s=q[i].a.size()-q[j].a.size();

                    if(q[j].a==q[i].a.substr(s,(int)q[j].a.size()))
                    {
//                        cout<<q[i].a.substr(s,(int)q[j].a.size())<<endl;
//                        cout<<q[j].a<<endl;
                        G[j].push_back(i);

                    }
                }
                else  if(q[i].a.size()==q[j].a.size())
                {
//                    int s=q[j].a.size();
                    if(q[j].a==q[i].a)
                    {
                        G[j].push_back(i);
                        G[i].push_back(j);
                    }
                }
                else
                {
                    int s=q[j].a.size()-q[i].a.size();
                    if(q[j].a.substr(s,q[i].a.size())==q[i].a)
                    {
                        G[i].push_back(j);

                    }
                }
            }
        }
//        for(int i=1; i<=n; i++)
//        {
//            for(int j=0; j<(int)G[i].size(); j++)
//            {
//                cout<<G[i][j]<<" ";
//            }
//            cout<<endl;
//
//        }
        int m;
        scanf("%d",&m);
        while(m--)
        {
            int e;
            scanf("%d",&e);
            if(e&1)
            {
                int q1,q2;
                scanf("%d %d",&q1,&q2);
                q[q1].b=q2;
            }
            else
            {
                int q1;
                scanf("%d",&q1);
                int sum=1;
                for(int i=0; i<(int)G[q1].size(); i++)
                {
                    if(q[G[q1][i]].b<=q[q1].b)
                        sum++;
                }
                printf("%d\n",sum);
            }
        }
    }
}