HDU 1528 Card Game cheater (minimum coverage)

#include "stdio.h" #include "string.h" #define N 52int map[n][n],v[n],link[n];int a[n],b[n],t,n;int dfs (int k) {int i;            for (i=1;i<=n;i++) {if (Map[k][i]&&!v[i]) {v[i]=1; if (link[i]==0| |                DFS (Link[i])) {link[i]=k;            return 1; }}} return 0;}    int main () {char c,d,e;    int I,j,vv,flag,ans;        int ap,bp;    scanf ("%d", &t);        while (t--) {scanf ("%d", &n);        GetChar ();        flag=1;        Ap=bp=0;            for (i=0;i<2*n;i++) {scanf ("%c%c%c", &c,&d,&e);            if (c>= ' 0 ' &&c<= ' 9 ') vv=c-' 0 ';            if (c== ' T ') vv=10;            if (c== ' J ') vv=11;            if (c== ' Q ') vv=12;            if (c== ' K ') vv=13;            if (c== ' A ') vv=14;            vv*=100;            if (d== ' H ') vv+=4;            if (d== ' S ') vv+=3;            if (d== ' d ') vv+=2;            if (d== ' C ') vv+=1;           if (flag)     A[++AP]=VV;            else B[++BP]=VV;        if (e== ' \ n ') flag=0;        } memset (map,0,sizeof (map));                    for (i=1;i<=n;i++) {for (j=1;j<=n;j++) {if (A[i]<b[j])            Map[j][i]=1;        }} memset (Link,0,sizeof (link));        ans=0;            for (i=1;i<=n;i++) {memset (v,0,sizeof (v));        if (Dfs (i)) ans++;    } printf ("%d\n", ans); } return 0;}

给定A,B两组牌,让找出B最多能的多少分 首先对牌的字符串全部转化成十进制数,然后以数字为点, B集合元素相对于A集合数字的大小关系为边,显然是典型的最小顶点覆盖问题 

HDU 1528 Card Game cheater (minimum coverage)