HDU1528Card Game Cheater（最大匹配）

来源：互联网发布：软件求职自我介绍编辑：程序博客网时间：2024/06/06 09:28

题意：给出 Adam和Eva二人的牌，游戏规则是，每个人轮流每次选择出一张牌，牌的数字大的赢，如果，数字一样按牌的花色H>S>D>C决定谁赢。
思路：按照牌的大小关系，建立二分图的A,B两个集合，EVA大于Adam的牌，就连上边。

#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<set>#include<map>#include<string>#include<cstring>#include<stack>#include<queue>#include<vector>#include<limits>#include<ctime>#include<cassert>#include<cstdlib>#define lson (rt<<1),L,M#define rson (rt<<1|1),M+1,R#define M ((L+R)>>1)#define cl(a,b) memset(a,b,sizeof(a));#define LL long long#define P pair<int,int>#define X first#define Y second#define pb push_back#define fread(a) freopen(a,"r",stdin);#define fwrite(a) freopen(a,"w",stdout);using namespace std;const int maxn=505;const int inf=999999;int x[maxn],y[maxn];void read_x(int n){//读入    for(int i=0;i<n;i++){        char op[3];        scanf("%s",op);        int num=0;        if(op[0]>='2'&&op[0]<='9')            num=op[0]-'0';        else if(op[0]=='T')            num=10;        else if(op[0]=='J')            num=11;        else if(op[0]=='Q')            num=12;        else if(op[0]=='K')            num=13;        else            num=14;        if(op[1]=='H')            num=num*10+4;        else if(op[1]=='S')            num=num*10+3;        else if(op[1]=='D')            num=num*10+2;        else            num=num*10+1;        x[i]=num;    }}void read_y(int n){    for(int i=0;i<n;i++){        char op[3];        scanf("%s",op);        int num=0;        if(op[0]>='2'&&op[0]<='9')            num=op[0]-'0';        else if(op[0]=='T')            num=10;        else if(op[0]=='J')            num=11;        else if(op[0]=='Q')            num=12;        else if(op[0]=='K')            num=13;        else            num=14;        if(op[1]=='H')            num=num*10+4;        else if(op[1]=='S')            num=num*10+3;        else if(op[1]=='D')            num=num*10+2;        else            num=num*10+1;        y[i]=num;    }    reverse(y,y+n);}vector<int> G[maxn];void build_map(int n){    for(int i=0;i<n;i++){       // printf("x = %d,y = %d\n",x[i],y[i]);        for(int j=0;j<n;j++)        if(x[i]<=y[j]){            G[i].pb(j);//满足关系，连上边        }    }}int matching[maxn];bool vis[maxn];int num;bool dfs(int u){    int N=G[u].size();    for(int i=0;i<N;i++){        int v=G[u][i];        if(vis[v])continue;        vis[v]=true;        if(matching[v]==-1||dfs(matching[v])){            matching[v]=u;            return true;        }    }    return false;}int hungar(){    int ans=0;    cl(matching,-1);    for(int i=0;i<num;i++){        cl(vis,false);        if(dfs(i))ans++;    }    return ans;}int main(){    int T;    scanf("%d",&T);    while(T--){        int n;        scanf("%d",&n);        for(int i=0;i<maxn;i++)G[i].clear();        read_x(n);        read_y(n);        build_map(n);        num=n;        int ans=hungar();        printf("%d\n",ans);    }    return 0;}

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