题目描述:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=26746
/*
solution:
节点表示插头类型,边表示转换器,然后使用floyd算法,计算出任意一种插头类型能否转换成另外一种插头类型。
额外添加一个源点s,从s到设备device[i]连接一条容量为1的边,再额外加一个汇点t,从插座target[i]到t连接一条
容量为1的边。然后只要device[i]能够转换成target[i]就在两者间添加一条容量为INF的边,表示允许任意多设备从
device[i]转换成target[i]。最后求s-t最大流,m减去最大流就是所要求的答案。
note:
1.关于EdmondKarp算法一般直接套模板即可。为了便于使用直接将其封装在结构体中
date:
2016/4/24
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
using namespace std;
const int maxn = 400 + 5;
const int maxm = 400 + 5;
const int maxk = 400 + 5;
const int INF = 100000000;
int n, m, k;
int target[maxn], device[maxm], d[maxn][maxn];
vector<string> typeVec;
struct Edge {
int from, to, cap, flow;
Edge(int u, int v, int c, int f):from(u),to(v),cap(c),flow(f) {}
};
struct EdmondsKarp {
int n, m;
vector<Edge> edges; // 边数的两倍
vector<int> G[maxn]; // 邻接表,G[i][j]表示结点i的第j条边在e数组中的序号
int a[maxn]; // 当起点到i的可改进量
int p[maxn]; // 最短路树上p的入弧编号
void init(int n) {
for(int i = 0; i < n; i++) G[i].clear();
edges.clear();
}
void addEdge(int from, int to, int cap) {
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, 0, 0));
m = edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
int maxflow(int s, int t) {
int flow = 0;
for(;;) {
memset(a, 0, sizeof(a));
queue<int> Q;
Q.push(s);
a[s] = INF;
while(!Q.empty()) {
int x = Q.front(); Q.pop();
for(int i = 0; i < G[x].size(); i++) {
Edge& e = edges[G[x][i]];
if(!a[e.to] && e.cap > e.flow) {
p[e.to] = G[x][i];
a[e.to] = min(a[x], e.cap-e.flow);
Q.push(e.to);
}
}
if(a[t]) break;
}
if(!a[t]) break;
for(int u = t; u != s; u = edges[p[u]].from) {
edges[p[u]].flow += a[t];
edges[p[u]^1].flow -= a[t];
}
flow += a[t];
}
return flow;
}
};
EdmondsKarp ek;
int ID(string str) {
int len = typeVec.size();
for(int i = 0; i < len; i++)
if(typeVec[i] == str) return i;
typeVec.push_back(str);
return len; //返回下标
}
void readInput() {
string s1, s2;
scanf("%d", &n); //插座
for(int i = 0; i < n; i++) {
cin >> s1;
target[i] = ID(s1);
}
scanf("%d", &m); //设备
for(int i = 0; i < m; i++) {
cin >> s1 >> s2;
device[i] = ID(s2);
}
scanf("%d", &k); //转换器
for(int i = 0; i < k; i++) {
cin >> s1 >> s2;
d[ID(s1)][ID(s2)] = 1;
}
}
void floyd() {
int v = typeVec.size();
for(int k = 0; k < v; k++)
for(int i = 0; i < v; i++)
for(int j = 0; j < v; j++)
d[i][j] = d[i][j] || (d[i][k] && d[k][j]);
}
void createGraph() {
int v =typeVec.size();
ek.init(v + 2);
//源点到设备
for(int i = 0; i < m; i++)
ek.addEdge(v, device[i], 1);
//插座到汇点
for(int i = 0; i < n; i++)
ek.addEdge(target[i], v + 1, 1);
for(int i = 0; i < m; i++)
for(int j = 0; j < n; j++)
if(d[device[i]][target[j]]) ek.addEdge(device[i], target[j], INF);
}
void init() {
typeVec.clear();
memset(target, 0, sizeof(target));
memset(device, 0, sizeof(device));
memset(d, 0, sizeof(d));
}
int main()
{
freopen("input.txt", "r", stdin);
int kase; scanf("%d", &kase);
getchar();
while(kase--) {
init();
readInput();
floyd();
createGraph();
int v = typeVec.size();
int r = ek.maxflow(v, v+1);
cout << m-r << "\n";
if(kase) cout << "\n";
}
return 0;
}