ZOJ 1157A Plug for UNIX(二分图最大匹配)

建图比较繁琐.

对于每一个adapter(a b) 连接 a -> b一条有向边.就是能把插在a的插座上的电器插到b插座上

然后对每一个插座i进行DFS,把能够到达的插座j标号,意思就是能把插在j的插座上的电器 插到i上.

最后匈牙利算法.

#include <iostream>
#include <cstdio>
#include <memory.h>
#include <vector>
#include <map>
#include <string>
using namespace std;
const int maxn = 600;

map<string, int> sToNum;
map<int,int> g2;
vector<int> g1[maxn];
map<int,map<int,int> > ada;
bool vis[maxn];
int fa[maxn],n, m, k, idx;
char buf1[maxn], buf2[maxn];
bool fg[maxn][maxn];
void dfs(int u, int o){
vis[u] = 1;
for (int i = 0; i < g1[u].size(); ++i){
if(!vis[g1[u][i]]){
dfs(g1[u][i],o);
ada[o][g1[u][i]] =1;
}
}
}
bool dfs2(int u){
for (int v = 1; v <= idx; ++v){
if(fg[u][v] && !vis[v]){
vis[v] = 1;
if (fa[v] ==0 || dfs2(fa[v])){
fa[v] = u;
return true;
}
}
}
return false;
}
int hungary(){
int ans = 0;
memset(fa, 0, sizeof(fa));
for (int i = 1; i <= n; ++i){
memset(vis, 0, sizeof(vis));
if(dfs2(i)){
ans++;
}
}
return ans;
}
void init(){
memset(fg, 0, sizeof(fg));
sToNum.clear();
ada.clear();
g2.clear();
for (int i = 0 ; i < maxn; ++i){
g1[i].clear();
}
}
int main(){
int T;
scanf("%d", &T);
while (T--){
init();
scanf("%d", &n);
int id = 1;
for (int i = 0; i < n; ++i){
scanf("%s", buf1);
sToNum[buf1] = id++;
}
scanf("%d", &m);
for (int i = 0; i < m; ++i){
scanf("%s %s", buf1, buf2);
int id1 = sToNum[buf1], id2 = sToNum[buf2];
if(id1 == 0){
id1 = sToNum[buf1] = id++;
}
if(id2 == 0){
id2 = sToNum[buf2] = id++;
}
g2[id1] = id2;
}
scanf("%d", &k);
for (int i = 0; i < k; ++i){
scanf("%s %s", buf1, buf2);
int id1 = sToNum[buf1], id2 = sToNum[buf2];
if(id1 == 0){
id1 = sToNum[buf1] = id++;
}
if(id2 == 0){
id2 = sToNum[buf2] = id++;
}
g1[id1].push_back(id2); //适配器a -> b 表示可以把 用a的插头的电器插在b插座上.
}
for (int i = 1; i <= id; ++i){
memset(vis, 0, sizeof(vis));
dfs(i,i);
}
map<int,int>::iterator it = g2.begin();
for (;it != g2.end(); ++it){
fg[it->first][it->second] = fg[it->second][it->first] =1;
map<int,int>::iterator it2 = ada[it->second].begin();
for (; it2 != ada[it->second].end(); ++it2){
fg[it->first][it2->first] = fg[it2->first][it->first] =1;
}
}
idx = id;
printf("%d\n", m - hungary());
if(T)printf("\n");
}
return 0;
}