题目是这样的,给定一些人喜欢某只猫或者狗,讨厌某只猫或者狗。求最多能够同时满足多少人的愿望?
题目很有意思。建模后就很简单了。
对于同一只猫或者狗,如果有一个讨厌,另一个人喜欢,那么这两个连一条边。最终,最大独立集数等于最大匹配数就可以了。。
Orz。
召唤代码君:
#include <iostream> #include <cstdio> #include <cstring> #include <vector> #define maxn 505 using namespace std; vector<int> likec[maxn],disc[maxn],liked[maxn],disd[maxn]; vector<int> v[maxn]; int n,m,p,ans,t1,t2; int f[maxn],tag[maxn]; char s1[maxn],s2[maxn]; void init() { ans=0; for (int i=1; i<=p; i++) v[i].clear(),f[i]=0,tag[i]=0; for (int i=1; i<=n; i++) likec[i].clear(),disc[i].clear(); for (int i=1; i<=m; i++) liked[i].clear(),disd[i].clear(); } int num(char S[]) { int cur=0; for (int i=0; S[i]; i++) cur=cur*10+S[i]-‘0‘; return cur; } int dfs(int cur,int T) { if (tag[cur]==T) return false; else tag[cur]=T; for (unsigned i=0; i<v[cur].size(); i++) { if (tag[v[cur][i]]==T) continue; if (f[v[cur][i]]==0 || dfs(f[v[cur][i]],T)) { f[v[cur][i]]=cur; f[cur]=v[cur][i]; return true; } } return false; } int main() { while (scanf("%d%d%d",&n,&m,&p)!=EOF) { init(); for (int i=1; i<=p; i++) { scanf("%s%s",s1,s2); t1=num(s1+1),t2=num(s2+1); if (s1[0]==‘C‘) likec[t1].push_back(i); else liked[t1].push_back(i); if (s2[0]==‘C‘) disc[t2].push_back(i); else disd[t2].push_back(i); } for (int i=1; i<=n; i++) for (unsigned x1=0; x1<likec[i].size(); x1++) for (unsigned x2=0; x2<disc[i].size(); x2++) { v[likec[i][x1]].push_back(disc[i][x2]); v[disc[i][x2]].push_back(likec[i][x1]); } for (int i=1; i<=m; i++) for (unsigned x1=0; x1<liked[i].size(); x1++) for (unsigned x2=0; x2<disd[i].size(); x2++) { v[liked[i][x1]].push_back(disd[i][x2]); v[disd[i][x2]].push_back(liked[i][x1]); } for (int i=1; i<=p; i++) { if (f[i]!=0) continue; if (dfs(i,i)) ans++; } printf("%d\n",p-ans); } return 0; }
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原文地址:http://www.cnblogs.com/Canon-CSU/p/3819181.html