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Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 15597 | Accepted: 5308 |
Description
Input
Output
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
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a,b) memset(a,b,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
const int big=50000;
int max(int a,int b) {return a>b?a:b;};
int min(int a,int b) {return a<b?a:b;};
struct edge{
int to,cap,rev;
};
vector<edge> G[2500];
map<string,int> mp;
int n,m,k,level[2500],iter[2500];
char s[50],a[50],b[50];
void add_edge(int u,int v,int cap)
{
G[u].push_back(edge{v,cap,G[v].size()});
G[v].push_back(edge{u,0,G[u].size()-1});
}
void bfs(int s)
{
queue<int> q;
q.push(s);
level[s]=1;
while(q.size())
{
int now=q.front();q.pop();
for(int i=0;i<G[now].size();i++)
if(G[now][i].cap>0)
{
edge e=G[now][i];
if(level[e.to]<0)
{
level[e.to]=level[now]+1;
q.push(e.to);
}
}
}
}
int dfs(int s,int t,int minn)
{
if(s==t)
return minn;
for(int &i=iter[s];i<G[s].size();i++)
{
edge &e=G[s][i];
if(level[e.to]>level[s]&&e.cap>0)
{
int k=dfs(e.to,t,min(minn,e.cap));
if(k>0)
{
e.cap-=k;
G[e.to][e.rev].cap+=k;
return k;
}
}
}
return 0;
}
int max_flow(int s,int t)
{
int ans=0,temp;
for(;;)
{
memset(level,-1,sizeof(level));
bfs(s);
if(level[t]<0)
return ans;
memset(iter,0,sizeof(iter));
while((temp=dfs(s,t,inf))>0)
ans+=temp;
}
return ans;
}
void init()
{
mp.clear();
for(int i=1;i<=2000;i++) G[i].clear();
}
void build()
{
int num=0;
for(int i=1;i<=n;i++)
{
int u;
scanf("%s",s);
mp[s]=++num;
add_edge(0,mp[s],1);
}
scanf("%d",&m);
for(int i=1;i<=m;i++)
{
scanf("%s %s",a,b);
mp[a]=++num;
if(!mp[b]) mp[b]=++num;
add_edge(mp[b],mp[a],1);
add_edge(mp[a],2000,1);
}
scanf("%d",&k);
for(int i=1;i<=k;i++)
{
scanf("%s %s",a,b);
if(!mp[a]) mp[a]=++num;
if(!mp[b]) mp[b]=++num;
add_edge(mp[b],mp[a],inf);
}
}
void solve()
{
printf("%d\n",m-max_flow(0,2000));
}
int main()
{
while(~scanf("%d",&n))
{
init();
build();
solve();
}
return 0;
}
分析:出的很渣的一道题,明明是道最大流裸题,结果题目讲的半天都含糊不清有歧义,,
最大流裸题,,唯一有收获的是map的使用map<string,int> mp保存字符串非常方便,直接scanf("%s",s) 然后mp[s]就可以读出mp中与s相同的字符串的个数
然后mp[]
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原文地址:http://www.cnblogs.com/smilesundream/p/5509608.html