hdu Dota all stars

Dota all stars

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 81    Accepted Submission(s): 35

Problem Description

Dota as a popular computer game has serval years. Lots of Dota fans know that the equipment in the game is very important. To get a excellent equipment may be pass many processes.  Look the picture, you may see that in order to get the finally one must get the other two first. But this two may be can not get directly. 　　　　　 　　　　　　　　　　　　　　　 
Now give the primary equipments which you can buy it by money directly in the shop. And the numbers of equipments you already have. You also know all the formulas how to get a new better equipment by primary equipments. Last give the numbers of equipments you want, tell us how much extra money you need.

 

Input

The input contains multiple test cases. First part a integer n1 expressing the numbers of primary equipments you can buy in the shop. Next n1 lines, each line form as S V,S is the name of equipment and V meaning you should cost V money to buy one equipments S. Second part a integer n2 expressing the numbers of kinds equipments you have. Next n2 lines, each line form as S,M, meaning the numbers of equipment S you have is M. Third part a integer n3 expressing the numbers of formulas. Next n3 lines, each line have a character '=', The left is the equipments you should use to get the right equipment. For example A + B + C = D, In order to get a D, you must use A,B,C,D.each one. Last part a integer n4, expressing n4 kinds of equipments you need. Than n4 lines, each line form as S,M. Meaning the numbers of equipment S you need is M.
You may sure the total kinds of equipment will not larger than 100. And the lengths of equipment name will less than 50.(a chinese words made up of two character )

 

Output

Output the extra money you need to achieve goal.

 

Sample Input

4
欢欣之刃 100
敏捷之靴 20
半兽人之斧 100
力量腰带 50
2
散华 1
力量腰带 2
3
欢欣之刃 + 敏捷之靴 = 散华
半兽人之斧 + 力量腰带 = 夜叉
散华 + 夜叉 = 散夜对剑
1
散夜对剑 2

5
鹰角弓 3200
短棍 1100
攻击之爪 950
阔剑 1200
恶魔刀锋 2600
0
3
鹰角弓 + 短棍 = 蝴蝶
攻击之爪 + 阔剑 = 水晶剑
水晶剑 + 恶魔刀锋 = 大炮
2
蝴蝶 1
大炮 1

 

Sample Output

320
9050

 

Author

yifenfei

 

Source

2009浙江大学计算机研考复试（机试部分）——全真模拟

 

Recommend

lcy

分析：很典型的dfs，本题难点在于用STL中的map来建立字符串与整数的对应。

#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<map>
using namespace std;
char s[55];
int hc[110][110];
int k[110];
int have[110];
int price[110];
int temp[110];
int money, flag,mo;
void dfs(int want)
{
   // printf("want=%d\n",want);
    if(have[want])
    {
    //    printf("have want=%d",want);
        flag=0;
        have[want]--;
    }
    else if(k[want])
    {
        int i;
        for(i=1;i<=k[want];++i)
            dfs(hc[want][i]);
    }
    else
    {
        money+=price[want];
    } 
}
int main() {
    int n1, n2, n3, n4, i;
    int order, cnt,t,n;
    while (scanf("%d", &n1) != EOF) {
        map<string, int> mymap;
        memset(k, 0, sizeof (k));
        money = order = 0;
        memset(have, 0, sizeof (have));
        memset(k, 0, sizeof (k));
        for (i = 1; i <= n1; ++i) {
            scanf("%s", s);
            if (!mymap[s])
                mymap[s] = ++order;
            scanf("%d", &price[mymap[s]]);
        }
        scanf("%d", &n2);
        for (i = 1; i <= n2; ++i) {
            scanf("%s", s);
            if (!mymap[s])
                mymap[s] = ++order;
            scanf("%d", &have[mymap[s]]);
        }
        scanf("%d", &n3);
        for (i = 1; i <= n3; ++i) {
            cnt = 0;
            while (1) {
                scanf("%s", s);
                if (!mymap[s])
                    mymap[s] = ++order;
                temp[++cnt] = mymap[s];
                scanf("%s", s);
                if (s[0] == '=')
                    break;
            }
            scanf("%s", s);
            if (!mymap[s])
                mymap[s] = ++order;
            k[mymap[s]] = cnt;
            while (cnt) {
                hc[mymap[s]][cnt] = temp[cnt];
                cnt--;
            }
        }
        scanf("%d", &n4);
        for (i = 1; i <= n4; ++i) {
            scanf("%s", s);
            if (!mymap[s])
                mymap[s] = ++order;
            t=mymap[s];
            scanf("%d", &n);
            while(n--)
            {
                flag=1;
                mo=money;
                dfs(t);
           //     printf("n=%d money=%d\n",n,money);
                if(flag)
                {
                    money+=n*(money-mo);
                    break;
                }
            }
            
        }
        printf("%d\n",money);
        
    }
    return 0;
}