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题目链接

Solution UVA12168 Cat vs. Dog

题目大意:给定\(n\)个人,有\(c\)只猫和\(d\)只狗.每个人会喜欢一只猫/狗,并且讨厌一只狗/猫.求一种方案让尽可能多的人满意(喜欢的动物出现,讨厌的不出现)

分析:这不就是二分图最大点独立集的模板吗.如果
\(u\)喜欢的是\(v\)讨厌的或者\(u\)讨厌的是\(v\)喜欢的,就要连边\((u,v)\)因为只有喜欢不同动物的人才可能有连边,所以这个图是二分图

两个人之间没有连边就代表他们的要求可以同时满足(都满意),跑一次二分图最大点独立集即可

注意毒瘤的输入!!!以及连边的顺序(要么猫爱好者连狗爱好者,要么反之)

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int INF = 0x7fffffff;
const int maxn = 1024;
const int maxm = (1024 * 1024 + 2048) << 1;
struct Edge{
    int from,to,cap,flow;
    Edge() = default;
    Edge(int a,int b,int c,int d):from(a),to(b),cap(c),flow(d){}
}Edges[maxm];
int head[maxn],nxt[maxm],tot = 1;
inline void clear(){
    memset(head,0,sizeof(head));
    memset(nxt,0,sizeof(nxt));
    tot = 1;
}
inline void addedge(int from,int to,int cap){
    Edges[++tot] = Edge(from,to,cap,0);
    nxt[tot] = head[from];
    head[from] = tot;
    Edges[++tot] = Edge(to,from,0,0);
    nxt[tot] = head[to];
    head[to] = tot;
}
int d[maxn];
inline bool bfs(int s,int t){
    memset(d,-1,sizeof(d));
    d[s] = 0;
    queue<int> Q;
    Q.push(s);
    while(!Q.empty()){
        int u = Q.front();Q.pop();
        for(int i = head[u];i;i = nxt[i]){
            Edge &e = Edges[i];
            if(e.cap > e.flow && d[e.to] == -1){
                d[e.to] = d[u] + 1;
                Q.push(e.to);
            }
        }
    }
    return d[t] != -1;
}
int cur[maxn];
inline int dfs(int u,int a,int t){
    if(u == t || a == 0)return a;
    int ret = 0,f;
    for(int &i = cur[u];i;i = nxt[i]){
        Edge &e = Edges[i];
        if(d[u] + 1 == d[e.to] && (f = dfs(e.to,min(a,e.cap - e.flow),t)) > 0){
            ret += f;
            Edges[i].flow += f;
            Edges[i ^ 1].flow -= f;
            a -= f;
            if(a == 0)break;
        }
    }
    return ret;
}
inline int maxflow(int s,int t){
    int ret = 0;
    while(bfs(s,t)){
        memcpy(cur,head,sizeof(head));
        ret += dfs(s,0x7fffffff,t);
    }
    return ret;
}
struct Person{
    string like,hate;
}person[maxn];
int t,cat,dog,n,cnt;
inline void solve(){
    clear();
    cin >> cat >> dog >> n;
    for(int i = 1;i <= n;i++){
        cin >> person[i].like >> person[i].hate;
    }
    int s = n + n + 1,t = n + n + 2;
    for(int i = 1;i <= n;i++)
        for(int j = 1;j <= n;j++)
            if((person[i].like[0] != person[j].like[0]) && (person[i].like == person[j].hate || person[i].hate == person[j].like)){
                if(person[i].like[0] == 'C')addedge(i,j + n,1);
                else if(person[j].like[0] == 'C')addedge(j,i + n,1);
            }
    for(int i = 1;i <= n;i++)
        if(person[i].like[0] == 'C')
            addedge(s,i,1);
    for(int i = 1;i <= n;i++)
        if(person[i].like[0] == 'D')
            addedge(i + n,t,1);
    cout << n - maxflow(s,t) << endl;
}
int main(){
#ifdef LOCAL
    freopen("fafa.in","r",stdin);
    freopen("fafa.out","w",stdout);
#else
    ios::sync_with_stdio(false);
#endif
    cin >> t;
    while(t--)
        solve();
    return 0;
}

题解 UVA12168 【Cat vs. Dog】

标签：ifd   emc   queue   open   int   out   name   while   https   

原文地址：https://www.cnblogs.com/colazcy/p/11515017.html