題目鏈接

Description

You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an international mandate to make the free flow of information and ideas on the Internet as cumbersome and bureaucratic as possible.
Since the room was designed to accommodate reporters and journalists from around the world, it is equipped with electrical receptacles to suit the different shapes of plugs and voltages used by appliances in all of the countries that existed when the room was built. Unfortunately, the room was built many years ago when reporters used very few electric and electronic devices and is equipped with only one receptacle of each type. These days, like everyone else, reporters require many such devices to do their jobs: laptops, cell phones, tape recorders, pagers, coffee pots, microwave ovens, blow dryers, curling
irons, tooth brushes, etc. Naturally, many of these devices can operate on batteries, but since the meeting is likely to be long and tedious, you want to be able to plug in as many as you can.
Before the meeting begins, you gather up all the devices that the reporters would like to use, and attempt to set them up. You notice that some of the devices use plugs for which there is no receptacle. You wonder if these devices are from countries that didn’t exist when the room was built. For some receptacles, there are several devices that use the corresponding plug. For other receptacles, there are no devices that use the corresponding plug.
In order to try to solve the problem you visit a nearby parts supply store. The store sells adapters that allow one type of plug to be used in a different type of outlet. Moreover, adapters are allowed to be plugged into other adapters. The store does not have adapters for all possible combinations of plugs and receptacles, but there is essentially an unlimited supply of the ones they do have.

Input

The input will consist of one case. The first line contains a single positive integer n (1 <= n <= 100) indicating the number of receptacles in the room. The next n lines list the receptacle types found in the room. Each receptacle type consists of a string of at most 24 alphanumeric characters. The next line contains a single positive integer m (1 <= m <= 100) indicating the number of devices you would like to plug in. Each of the next m lines lists the name of a device followed by the type of plug it uses (which is identical to the type of receptacle it requires). A device name is a string of at most 24 alphanumeric
characters. No two devices will have exactly the same name. The plug type is separated from the device name by a space. The next line contains a single positive integer k (1 <= k <= 100) indicating the number of different varieties of adapters that are available. Each of the next k lines describes a variety of adapter, giving the type of receptacle provided by the adapter, followed by a space, followed by the type of plug.

Output

A line containing a single non-negative integer indicating the smallest number of devices that cannot be plugged in.

Sample Input

4 A B C D 5 laptop B phone C pager B clock B comb X 3 B X X A X D

Sample Output

1

AC

• 建圖：

• 將源點和電器需要插座建邊，權值為需要的個數
• 將插座轉換器建邊，權值為INF（商店可以提供無數個）
• 將已有的插座和匯點相連，權值為該種類插座的數目

• 注意：

• 如果用鄰接矩陣寫，矩陣最小開到400，因為插座的種類最壞的情況是400種，已存在的插座100種，需要的插座100種，轉換器100 + 100 = 200 種，所以最多400種

• 代碼后面有極限數據

#include <iostream> #include <stdio.h> #include <map> #include <vector> #include <queue> #include <algorithm> #include <cmath> #define N 300005 #include <cstring> #define ll long long #define P pair<int, int> #define mk make_pair using namespace std;int pre[405], a[405][405]; bool vis[405]; int inf = 0x3f3f3f3f; int n, m, k; // EK最大流 bool bfs(int s, int e) {memset(pre, -1, sizeof(pre));memset(vis, false, sizeof(vis));vis[s] = true;pre[s] = s;queue<int> que;que.push(s);while (!que.empty()) {int t = que.front();que.pop();for (int i = 0; i <= e; ++i) {if (vis[i] || a[t][i] == 0) continue;vis[i] = true;pre[i] = t;if (i == e) return true;que.push(i);}}return false; } ll solve(int s, int e) {ll sum = 0;while (bfs(s, e)) {int MIN = inf;for (int i = e; i != s; i = pre[i]) {MIN = min(MIN, a[pre[i]][i]);}for (int i = e; i != s; i = pre[i]) {a[pre[i]][i] -= MIN;a[i][pre[i]] += MIN;}sum += MIN;}return sum; }int main() { #ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin); #endifios::sync_with_stdio(false);while (cin >> n) {memset(a, 0, sizeof(a));// 統計現有的插座種類和數目 map<string, int> mp1;// 統計電器插頭種類和數目 map<string, int> mp2;// 對所有插座和插頭進行標記 map<string, int> mp;int now = 1;// 讀入插座 for (int i = 0; i < n; ++i) {string s;cin >> s;mp1[s]++;if (mp[s] == 0) mp[s] = now++;}// 讀入插頭 cin >> m;for (int i = 0; i < m; ++i) {string s;cin >> s >> s;mp2[s]++; if (mp[s] == 0) mp[s] = now++;}// 將源點和電器需要的插座相連，權值為需要的個數 map<string, int>::iterator it;for (it = mp2.begin(); it != mp2.end(); ++it) {a[0][mp[it->first]] = it->second;}// 讀入轉換器 cin >> k;for (int i = 0; i < k; ++i) {string s1, s2;cin >> s1 >> s2;if (mp[s1] == 0) mp[s1] = now++;if (mp[s2] == 0) mp[s2] = now++;// 因為商店可以提供無數個，權值設為 inf a[mp[s1]][mp[s2]] = inf;}// 將現有的插座和匯點相連 for (it = mp1.begin(); it != mp1.end(); ++it) {a[mp[it->first]][401] = it->second;}// 插座和插頭最多400種，已有的100種，需要的100種，轉換器100 + 100 ll ans = solve(0, 401);printf("%d\n", m - ans);}return 0; }

• 數據

100 a0 a1 a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 a12 a13 a14 a15 a16 a17 a18 a19 a20 a21 a22 a23 a24 a25 a26 a27 a28 a29 a30 a31 a32 a33 a34 a35 a36 a37 a38 a39 a40 a41 a42 a43 a44 a45 a46 a47 a48 a49 a50 a51 a52 a53 a54 a55 a56 a57 a58 a59 a60 a61 a62 a63 a64 a65 a66 a67 a68 a69 a70 a71 a72 a73 a74 a75 a76 a77 a78 a79 a80 a81 a82 a83 a84 a85 a86 a87 a88 a89 a90 a91 a92 a93 a94 a95 a96 a97 a98 a99 100 b0 c0 b1 c1 b2 c2 b3 c3 b4 c4 b5 c5 b6 c6 b7 c7 b8 c8 b9 c9 b10 c10 b11 c11 b12 c12 b13 c13 b14 c14 b15 c15 b16 c16 b17 c17 b18 c18 b19 c19 b20 c20 b21 c21 b22 c22 b23 c23 b24 c24 b25 c25 b26 c26 b27 c27 b28 c28 b29 c29 b30 c30 b31 c31 b32 c32 b33 c33 b34 c34 b35 c35 b36 c36 b37 c37 b38 c38 b39 c39 b40 c40 b41 c41 b42 c42 b43 c43 b44 c44 b45 c45 b46 c46 b47 c47 b48 c48 b49 c49 b50 c50 b51 c51 b52 c52 b53 c53 b54 c54 b55 c55 b56 c56 b57 c57 b58 c58 b59 c59 b60 c60 b61 c61 b62 c62 b63 c63 b64 c64 b65 c65 b66 c66 b67 c67 b68 c68 b69 c69 b70 c70 b71 c71 b72 c72 b73 c73 b74 c74 b75 c75 b76 c76 b77 c77 b78 c78 b79 c79 b80 c80 b81 c81 b82 c82 b83 c83 b84 c84 b85 c85 b86 c86 b87 c87 b88 c88 b89 c89 b90 c90 b91 c91 b92 c92 b93 c93 b94 c94 b95 c95 b96 c96 b97 c97 b98 c98 b99 c99 100 d0 e0 d1 e1 d2 e2 d3 e3 d4 e4 d5 e5 d6 e6 d7 e7 d8 e8 d9 e9 d10 e10 d11 e11 d12 e12 d13 e13 d14 e14 d15 e15 d16 e16 d17 e17 d18 e18 d19 e19 d20 e20 d21 e21 d22 e22 d23 e23 d24 e24 d25 e25 d26 e26 d27 e27 d28 e28 d29 e29 d30 e30 d31 e31 d32 e32 d33 e33 d34 e34 d35 e35 d36 e36 d37 e37 d38 e38 d39 e39 d40 e40 d41 e41 d42 e42 d43 e43 d44 e44 d45 e45 d46 e46 d47 e47 d48 e48 d49 e49 d50 e50 d51 e51 d52 e52 d53 e53 d54 e54 d55 e55 d56 e56 d57 e57 d58 e58 d59 e59 d60 e60 d61 e61 d62 e62 d63 e63 d64 e64 d65 e65 d66 e66 d67 e67 d68 e68 d69 e69 d70 e70 d71 e71 d72 e72 d73 e73 d74 e74 d75 e75 d76 e76 d77 e77 d78 e78 d79 e79 d80 e80 d81 e81 d82 e82 d83 e83 d84 e84 d85 e85 d86 e86 d87 e87 d88 e88 d89 e89 d90 e90 d91 e91 d92 e92 d93 e93 d94 e94 d95 e95 d96 e96 d97 e97 d98 e98 d99 e99

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