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。
1 #include <cstdio> 2 #include <cstring> 3 #include <cmath> 4 #include <cstdlib> 5 #include <algorithm> 6 #include <string> 7 #include <iostream> 8 #include <stack> 9 #include <map> 10 #include <queue> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 #define N 100010 15 #define INF 0x3f3f3f3f 16 int s[] = {0, 16, 7, 8, 1, 1, 2, 3}; 17 int mp1[15], mp2[15]; 18 19 int main() 20 { 21 2, 3; 22 int t; 23 scanf("%d", &t); 24 while(t--) { 25 int red = 0, bla = 0; 26 memset(mp1, 0, sizeof(mp1)); 27 memset(mp2, 0, sizeof(mp2)); 28 int n; char str[3]; 29 for(int i = 1; i <= 2; i++) { 30 scanf("%d", &n); 31 if(i == 1) { 32 for(int j = 0; j < n; j++) { 33 scanf("%s", str); 34 red += s[str[0]-'A'+1]; 35 mp1[str[0]-'A'+1]++; 36 } 37 } else { 38 for(int j = 0; j < n; j++) { 39 scanf("%s", str); 40 bla += s[str[0]-'A'+1]; 41 mp2[str[0]-'A'+1]++; 42 } 43 } 44 } 45 if(mp1[2] == 0 || mp1[3] == 0) if(red > 0) red--; 46 if(mp2[2] == 0 || mp2[3] == 0) if(bla > 0) bla--; 47 if(red == bla) puts("tie"); 48 else if(red < bla) puts("black"); 49 else puts("red"); 50 } 51 return 0; 52 }
