HDU 4461：The Power of Xiangqi（水题）

http://acm.hdu.edu.cn/showproblem.php?pid=4461

题意：每个棋子有一个权值，给出红方的棋子情况，黑方的棋子情况，问谁能赢。

思路：注意“ if a player has no Ma or no Pao, or has neither, his total offense power will be decreased by one”这句话即可。

 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <cstdlib>
 #include <algorithm>
 #include <string>
 #include <iostream>
 #include <stack>
 #include <map>
 #include <queue>
 #include <set>
 using namespace std;
 typedef long long LL;
 #define N 100010
 #define INF 0x3f3f3f3f
 int s[] = {, , , , , , , };
 int mp1[], mp2[];

 int main()
 {
     , ;
     int t;
     scanf("%d", &t);
     while(t--) {
         int red = , bla = ;
         memset(mp1, , sizeof(mp1));
         memset(mp2, , sizeof(mp2));
         int n; char str[];
         for(int i = ; i <= ; i++) {
             scanf("%d", &n);
             if(i == ) {
                 for(int j = ; j < n; j++) {
                     scanf("%s", str);
                     red += s[str[]-'A'+];
                     mp1[str[]-'A'+]++;
                 }
             } else {
                 for(int j = ; j < n; j++) {
                     scanf("%s", str);
                     bla += s[str[]-'A'+];
                     mp2[str[]-'A'+]++;
                 }
             }
         }
         if(mp1[] ==  || mp1[] == ) if(red > ) red--;
         if(mp2[] ==  || mp2[] == ) if(bla > ) bla--;
         if(red == bla) puts("tie");
         else if(red < bla) puts("black");
         else puts("red");
     }
     return ;
 }