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[] = {0, 16, 7, 8, 1, 1, 2, 3};
int mp1[15], mp2[15];

int main()
{
    2, 3;
    int t;
    scanf("%d", &t);
    while(t--) {
        int red = 0, bla = 0;
        memset(mp1, 0, sizeof(mp1));
        memset(mp2, 0, sizeof(mp2));
        int n; char str[3];
        for(int i = 1; i <= 2; i++) {
            scanf("%d", &n);
            if(i == 1) {
                for(int j = 0; j < n; j++) {
                    scanf("%s", str);
                    red += s[str[0]-'A'+1];
                    mp1[str[0]-'A'+1]++;
                }
            } else {
                for(int j = 0; j < n; j++) {
                    scanf("%s", str);
                    bla += s[str[0]-'A'+1];
                    mp2[str[0]-'A'+1]++;
                }
            }
        }
        if(mp1[2] == 0 || mp1[3] == 0) if(red > 0) red--;
        if(mp2[2] == 0 || mp2[3] == 0) if(bla > 0) bla--;
        if(red == bla) puts("tie");
        else if(red < bla) puts("black");
        else puts("red");
    }
    return 0;
}