首页 > 代码库 > POJ 1087 A Plug for UNIX 会议室插座问题 构图+最大流
POJ 1087 A Plug for UNIX 会议室插座问题 构图+最大流
题目链接：POJ 1087 A Plug for UNIX
A Plug for UNIX
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 Source East Central North America 1999 |
分析：
发现基本构图好了，网络流的题目就很好解了。
（1）以0为源点，1为汇点，其他的插座还有设备都作为中间点
（2）会议室提供n个插座，从源点到每个插座连一条边，容量为1
（3）会议室有m个设备，从每个设备到汇点连一条边，容量为1
（4）每个设备使用一个插座，从相应插座到设备连一条边，容量为1
（5）有k中转接器，从插头到转接器提供插座类型连一条边，即前者可以转化为后者，容量为无穷，因为可以串联。
（6）求从源点到汇点最大流，及最多使用设备数目maxflow，最后结果为m-maxflow。
代码;
Dinic：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
using namespace std;
#define maxn 1010
#define INF 0x3f3f3f3f
struct Edge
{
int from, to, cap;
};
vector<Edge> EG;
vector<int> G[maxn];
int n, s, t, d[maxn], cur[maxn], mp[maxn][maxn];
char name[maxn][30];
int cnt;
void addEdge(int from, int to, int cap)
{
EG.push_back((Edge){from, to, cap});
EG.push_back((Edge){to, from, 0});
int x = EG.size();
G[from].push_back(x-2);
G[to].push_back(x-1);
}
bool bfs()
{
memset(d, -1, sizeof(d));
queue<int> q;
q.push(s);
d[s] = 0;
while(!q.empty())
{
int x = q.front();
q.pop();
for(int i = 0; i < G[x].size(); i++)
{
Edge& e = EG[G[x][i]];
if(d[e.to] == -1 && e.cap > 0)
{
d[e.to] = d[x]+1;
q.push(e.to);
}
}
}
return (d[t]!=-1);
}
int dfs(int x, int a)
{
if(x == t || a == 0) return a;
int flow = 0, f;
for(int& i = cur[x]; i < G[x].size(); i++)
{
Edge& e = EG[G[x][i]];
if(d[x]+1 == d[e.to] && (f = dfs(e.to, min(a, e.cap))) > 0)
{
e.cap -= f;
EG[G[x][i]^1].cap += f;
flow += f;
a -= f;
if(a == 0) break;
}
}
return flow;
}
int Dinic()
{
int ans = 0;
while(bfs())
{
memset(cur, 0, sizeof(cur));
ans += dfs(s, INF);
}
EG.clear();
for(int i = 0; i < n; ++i)
G[i].clear();
return ans;
}
int Find(char* str)
{
int i;
for(i = 2; i < cnt; ++i)
if(strcmp(name[i], str) == 0)
return i;
strcpy(name[i], str);
cnt++;
return i;
}
int main()
{
//freopen("poj_1087.txt", "r", stdin);
int m, k;
char str1[30], str2[30];
while(~scanf("%d", &n)) {
s = 0, t = 1;
cnt = 2; //源点和汇点占两个
for(int i = 0; i < n; i++) {
scanf("%s", str1);
strcpy(name[cnt], str1); //插座不会从父，直接插入
cnt++;
addEdge(0, i+2, 1); //建边，源点向每个插座连一条1的边
}
scanf("%d", &m);
for(int i = 0; i < m; i++) {
scanf("%s%s", str1, str2);
strcpy(name[cnt], str1); //设备也不会重复。直接插入
cnt++;
int u = Find(str2); //扎里插座可能重复，要查找
addEdge(u, cnt-1, 1); //建边，插座向设备连一条容量为1的边
addEdge(cnt-1, 1, 1); //建边，设备到汇点连一条边，容量为1
}
scanf("%d", &k);
for(int i = 0; i < k; i++) {
scanf("%s%s", str1, str2);
int u = Find(str1);
int v = Find(str2);
addEdge(v, u, INF); //建边，后者到前者容量为无穷
}
n = cnt;
int ans = Dinic();
printf("%d\n", m-ans);
}
return 0;
}
ISAP：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
using namespace std;
#define maxn 1010
#define INF 0x3f3f3f3f
struct Edge
{
int from, to, cap, flow;
};
char name[maxn][30];
vector<Edge> EG;
vector<int> G[maxn];
int n, s, t, d[maxn], cur[maxn], p[maxn], num[maxn], mp[maxn][maxn];
bool vis[maxn];
int cnt;
void addEdge(int from, int to, int cap)
{
EG.push_back((Edge){from, to, cap, 0});
EG.push_back((Edge){to, from, 0, 0});
int x = EG.size();
G[from].push_back(x-2);
G[to].push_back(x-1);
}
void bfs()
{
memset(vis, false, sizeof(vis));
queue<int> q;
vis[t] = true;
d[t] = 0;
q.push(t);
while(!q.empty()) {
int x = q.front();
q.pop();
for(int i = 0; i < G[x].size(); i++) {
Edge& e = EG[G[x][i]^1];
if(!vis[e.from] && e.cap > e.flow) {
vis[e.from] = true;
d[e.from] = d[x]+1;
q.push(e.from);
}
}
}
}
int augment()
{
int x = t, a = INF;
while(x != s) {
Edge& e = EG[p[x]];
a = min(a, e.cap-e.flow);
x = EG[p[x]].from;
}
x = t;
while(x != s) {
EG[p[x]].flow += a;
EG[p[x]^1].flow -= a;
x = EG[p[x]].from;
}
return a;
}
int ISAP()
{
int ans =0;
bfs();
memset(num, 0, sizeof(num));
for(int i = 0; i < n; i++)
num[d[i]]++;
int x = s;
memset(cur, 0, sizeof(cur));
while(d[s] < n) {
if(x == t) {
ans += augment();
x = s;
}
bool flag = false;
for(int i = cur[x]; i < G[x].size(); i++) {
Edge& e = EG[G[x][i]];
if(e.cap > e.flow && d[x] == d[e.to]+1) {
flag = true;
p[e.to] = G[x][i];
cur[x] = i;
x = e.to;
break;
}
}
if(!flag) {
int m = n-1;
for(int i = 0; i < G[x].size(); i++) {
Edge& e = EG[G[x][i]];
if(e.cap > e.flow)
m = min(m, d[e.to]);
}
if(--num[d[x]] == 0) break;
num[d[x] = m+1]++;
cur[x] = 0;
if(x != s)
x = EG[p[x]].from;
}
}
EG.clear();
for(int i = 0; i < n; ++i)
G[i].clear();
return ans;
}
int Find(char* str)
{
int i;
for(i = 2; i < cnt; ++i)
if(strcmp(name[i], str) == 0)
return i;
strcpy(name[i], str);
cnt++;
return i;
}
int main()
{
//freopen("poj_1087.txt", "r", stdin);
int m, k;
char str1[30], str2[30];
while(~scanf("%d", &n)) {
s = 0, t = 1;
cnt = 2;
for(int i = 0; i < n; i++) {
scanf("%s", str1);
strcpy(name[cnt], str1);
cnt++;
addEdge(0, i+2, 1);
}
scanf("%d", &m);
for(int i = 0; i < m; i++) {
scanf("%s%s", str1, str2);
strcpy(name[cnt], str1);
cnt++;
int u = Find(str2);
addEdge(u, cnt-1, 1);
addEdge(cnt-1, 1, 1);
}
scanf("%d", &k);
for(int i = 0; i < k; i++) {
scanf("%s%s", str1, str2);
int u = Find(str1);
int v = Find(str2);
addEdge(v, u, INF);
}
n = cnt;
int ans = ISAP();
printf("%d\n", m-ans);
}
return 0;
}
不知为何，总感觉我的Dinic比ISAP要快，基本每次都是，不应该啊。
POJ 1087 A Plug for UNIX 会议室插座问题 构图+最大流
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