題干：

Berland shop sells?nn?kinds of juices. Each juice has its price?cici. Each juice includes some set of vitamins in it. There are three types of vitamins: vitamin "A", vitamin "B" and vitamin "C". Each juice can contain one, two or all three types of vitamins in it.

Petya knows that he needs all three types of vitamins to stay healthy. What is the minimum total price of juices that Petya has to buy to obtain all three vitamins? Petya obtains some vitamin if he buys at least one juice containing it and drinks it.

Input

The first line contains a single integer?nn?(1≤n≤1000)(1≤n≤1000)?— the number of juices.

Each of the next?nn?lines contains an integer?cici?(1≤ci≤100000)(1≤ci≤100000)?and a string?sisi?— the price of the?ii-th juice and the vitamins it contains. String?sisi?contains from?11to?33?characters, and the only possible characters are "A", "B" and "C". It is guaranteed that each letter appears no more than once in each string?sisi. The order of letters in strings?sisi?is arbitrary.

Output

Print?-1?if there is no way to obtain all three vitamins. Otherwise print the minimum total price of juices that Petya has to buy to obtain all three vitamins.

Examples

Input

4 5 C 6 B 16 BAC 4 A

Output

15

Input

2 10 AB 15 BA

Output

-1

Input

5 10 A 9 BC 11 CA 4 A 5 B

Output

13

Input

6 100 A 355 BCA 150 BC 160 AC 180 B 190 CA

Output

250

Input

2 5 BA 11 CB

Output

16

Note

In the first example Petya buys the first, the second and the fourth juice. He spends?5+6+4=155+6+4=15?and obtains all three vitamins. He can also buy just the third juice and obtain three vitamins, but its cost is?1616, which isn't optimal.

In the second example Petya can't obtain all three vitamins, as no juice contains vitamin "C".

解題報告：

? ?別忘了考慮? ‘AB’和‘BC’這類的三種情況，‘AB’和‘ABC’這種就不需要了，因為還不如直接'ABC'呢。

AC代碼：

#include<bits/stdc++.h>using namespace std; int n,m; map<string,int> mp; int main() {int n;string s;int tmp;cin>>n; for(int i = 1; i<=n; i++) {cin>>tmp>>s;sort(s.begin(),s.end());if(mp.find(s) == mp.end()) mp[s] = tmp;else mp[s] = min(mp[s],tmp);}int minn = 0x3f3f3f3f;if(mp["A"]>0 && mp["B"]>0 && mp["C"]>0) minn = min(minn,mp["A"] + mp["B"] + mp["C"]);if(mp["AB"]>0 && mp["C"]>0) minn = min(minn,mp["AB"] + mp["C"]);if(mp["A"]>0 && mp["BC"]>0) minn = min(minn,mp["A"] + mp["BC"]);if(mp["AC"]>0 && mp["B"]>0) minn = min(minn,mp["AC"] + mp["B"]);if(mp["ABC"]>0) minn = min(minn,mp["ABC"]);if(mp["AB"] && mp["AC"]) minn = min(minn,mp["AB"] + mp["AC"]); if(mp["AB"] && mp["BC"]) minn = min(minn,mp["AB"] + mp["BC"]); if(mp["AC"] && mp["BC"]) minn = min(minn,mp["AC"] + mp["BC"]); if(minn == 0x3f3f3f3f) puts("-1");else printf("%d\n",minn); return 0; }

其他的一堆AC代碼：（有用dp的，有無限更新的）

#include <bits/stdc++.h>/* unsigned seed1 = std::chrono::system_clock::now().time_since_epoch().count(); mt19937 g1.seed(seed1);ios_base::sync_with_stdio(false); cin.tie(NULL); */ using namespace std;int dp[8]; int n;char buf[10];int main() {scanf("%d", &n);for(int i = 1; i < 8; i++) {dp[i] = 1e9;}while(n--) {int s;scanf("%d %s", &s, buf);int m = 0;for(int i = 0; i < strlen(buf); i++) {m |= 1 << (buf[i] - 'A');}for(int i = 0; i < 8; i++) {dp[i|m] = min(dp[i|m], dp[i] + s);}}if(dp[7] == 1e9) dp[7] = -1;printf("%d\n", dp[7]); }#include<bits/stdc++.h> using namespace std; typedef long long ll;int N; int m[8];int main() {ios_base::sync_with_stdio(0);cin.tie(NULL);for (int i = 0; i < 8; i++)m[i] = 1e6;cin >> N;for (int i = 0; i < N; i++) {int c;string s;cin >> c >> s;int cc = 0;for (int j = 0; j < s.length(); j++)cc += (1 << (s[j] - 'A'));m[cc] = min (m[cc], c);}int mans = 1e9;for (int i = 0; i < 256; i++) {int b = 0, cc = 0;for (int j = 0; j < 8; j++) {if (i & (1 << j)) {b |= j;cc += m[j];}}if (b == 7)mans = min (mans, cc);}if (mans >= 1e6)cout << "-1\n";elsecout << mans << "\n"; }#include <bits/stdc++.h> using namespace std; typedef long long LL; const int inf=1e9; int n, c, bst[8]; char buf[8]; int main() {for(int i=1; i<8; i++) bst[i]=inf;scanf("%d", &n);while(n--) {scanf("%d%s", &c, buf);int msk=0;for(int i=0; buf[i]; i++)msk|=(1<<(buf[i]-'A'));bst[msk]=min(bst[msk], c);}for(int rnd=0; rnd<514; rnd++)for(int i=0; i<8; i++)for(int j=0; j<8; j++)bst[i|j]=min(bst[i|j], bst[i]+bst[j]);if(bst[7] == inf) bst[7]=-1;cout<<bst[7]<<endl; }#include <bits/stdc++.h> using namespace std;const int INF = 123456789;int p[8];int main () {int n;cin >> n;for (int i=0; i<8; ++i) {p[i] = INF;}while (n--) {int x;string s;cin >> x >> s;int mask = 0;for (char i:s) {mask |= (1<<(i-'A'));}p[mask] = min(p[mask],x);}int cost = INF;for (int i=0; i<8; ++i) {if (i == 7) {cost = min(cost,p[i]);}}for (int i=0; i<8; ++i) {for (int j=0; j<8; ++j) {if ((i|j) == 7) {cost = min(cost,p[i]+p[j]);}}}for (int i=0; i<8; ++i) {for (int j=0; j<8; ++j) {for (int k=0; k<8; ++k) {if ((i|j|k) == 7) {cost = min(cost,p[i]+p[j]+p[k]);}}}}if (cost == INF) cout << -1 << endl;else cout << cost << endl; }

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