题意：两个人A和B在打牌，只有题目给出的几种牌能出若A第一次出牌B压不住或者A一次就把牌出完了，那么A赢，输出Yes，否则若A牌没出完而且被B压住了，那么A输，输出No。
解法：知道规则，看清题目，搞清有哪些Trick，就可以直接模拟搞了。详见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
#define N 100102
#define M 22
char s1[24],s2[24];
int A[24],B[24];
int cnt1[20],cnt2[20];
int main()
{
int n,i,j;
int n1,n2;
scanf("%d",&n);
while(n--)
{
priority_queue<int> Single,Pair,Tri,Four,Nuke;
priority_queue<int> Single2,Pair2,Tri2,Four2,Nuke2;
scanf("%s",s1);
scanf("%s",s2);
n1 = strlen(s1);
n2 = strlen(s2);
for(i=0;i<n1;i++)
{
if(s1[i] >= ‘3‘ && s1[i] <= ‘9‘)
A[i] = s1[i]-‘0‘;
else if(s1[i] == ‘T‘)
A[i] = 10;
else if(s1[i] == ‘J‘)
A[i] = 11;
else if(s1[i] == ‘Q‘)
A[i] = 12;
else if(s1[i] == ‘K‘)
A[i] = 13;
else if(s1[i] == ‘A‘)
A[i] = 14;
else if(s1[i] == ‘2‘)
A[i] = 15;
else if(s1[i] == ‘X‘)
A[i] = 16;
else if(s1[i] == ‘Y‘)
A[i] = 17;
}
for(i=0;i<n2;i++)
{
if(s2[i] >= ‘3‘ && s2[i] <= ‘9‘)
B[i] = s2[i]-‘0‘;
else if(s2[i] == ‘T‘)
B[i] = 10;
else if(s2[i] == ‘J‘)
B[i] = 11;
else if(s2[i] == ‘Q‘)
B[i] = 12;
else if(s2[i] == ‘K‘)
B[i] = 13;
else if(s2[i] == ‘A‘)
B[i] = 14;
else if(s2[i] == ‘2‘)
B[i] = 15;
else if(s2[i] == ‘X‘)
B[i] = 16;
else if(s2[i] == ‘Y‘)
B[i] = 17;
}
sort(A,A+n1);
sort(B,B+n2);
memset(cnt1,0,sizeof(cnt1));
memset(cnt2,0,sizeof(cnt2));
for(i=0;i<n1;i++) //计算A各种牌的个数
cnt1[A[i]]++;
for(i=0;i<n2;i++) //计算B各种牌的个数
cnt2[B[i]]++;
for(i=3;i<=15;i++)
{
if(cnt1[i] == 4) //出现四个，可以做四个出，可以做三个出，也可以做两个或一个出
Four.push(i),Tri.push(i),Pair.push(i),Single.push(i);
else if(cnt1[i] == 3)
Tri.push(i),Pair.push(i),Single.push(i);
else if(cnt1[i] == 2)
Pair.push(i),Single.push(i);
else if(cnt1[i] == 1)
Single.push(i);
}
for(i=3;i<=15;i++)
{
if(cnt2[i] == 4)
Four2.push(i),Tri2.push(i),Pair2.push(i),Single2.push(i);
else if(cnt2[i] == 3)
Tri2.push(i),Pair2.push(i),Single2.push(i);
else if(cnt2[i] == 2)
Pair2.push(i),Single2.push(i);
else if(cnt2[i] == 1)
Single2.push(i);
}
if(cnt1[16]) //有王，可以做单个出
Nuke.push(16),Single.push(16);
if(cnt1[17])
Nuke.push(17),Single.push(17);
if(cnt2[16])
Nuke2.push(16),Single2.push(16);
if(cnt2[17])
Nuke2.push(17),Single2.push(17);
if(Nuke.size() >= 2) //双王，直接赢
{
puts("Yes");
continue;
}
//-------------------------------下面判断能否一次出完
if(n1 == 1)
{
puts("Yes");
continue;
}
if(n1 == 2)
{
if(A[0] == A[1])
{
puts("Yes");
continue;
}
}
if(n1 == 3)
{
if(A[0] == A[1] && A[1] == A[2])
{
puts("Yes");
continue;
}
}
if(n1 == 4)
{
if(A[0] == A[1] && A[1] == A[2] && A[2] == A[3])
{
puts("Yes");
continue;
}
if(A[0] != A[1] && A[1] == A[2] && A[2] == A[3])
{
puts("Yes");
continue;
}
if(A[2] != A[3] && A[0] == A[1] && A[1] == A[2])
{
puts("Yes");
continue;
}
}
if(n1 == 5)
{
if(A[0] == A[1] && A[1] != A[2] && A[2] == A[3] && A[3] == A[4])
{
puts("Yes");
continue;
}
if(A[3] == A[4] && A[2] != A[3] && A[0] == A[1] && A[1] == A[2])
{
puts("Yes");
continue;
}
}
if(n1 == 6)
{
int tag = 0;
for(i=0;i<=2;i++)
{
if(A[i] == A[i+1] && A[i+1] == A[i+2] && A[i+2] == A[i+3])
{
tag = 1;
break;
}
}
if(tag)
{
puts("Yes");
continue;
}
}
//-----------------------------如果不能一次出完
if(Nuke2.size() >= 2) //对方有双王，必输
{
puts("No");
continue;
}
if(!Nuke.empty() && Nuke2.empty()) //A有王，B没王
{
puts("Yes");
continue;
}
if(!Nuke.empty() && !Nuke2.empty()) //都有王，看谁的大，如果A小，则不选择出王，继续
{
if(Nuke.top() > Nuke2.top())
{
puts("Yes");
continue;
}
}
if(Four.empty() && !Four2.empty()) //炸弹，如果不能一次出完又没炸弹，那么必会被炸，输
{
puts("No");
continue;
}
if(!Four.empty() && Four2.empty()) //有炸弹出炸弹
{
puts("Yes");
continue;
}
if(!Four.empty() && !Four2.empty()) //都有炸弹，A的如果小，因为不能一次出完，必输
{
if(Four.top() >= Four2.top())
{
puts("Yes");
continue;
}
else
{
puts("No");
continue;
}
}
if(!Tri.empty() && Tri2.empty()) //三个的情况
{
puts("Yes");
continue;
}
if(!Tri.empty() && !Tri2.empty())
{
if(Tri.top() >= Tri2.top())
{
puts("Yes");
continue;
}
else if(n1 >= 4 && n2 <= 3) //A有的带，B没得带
{
puts("Yes");
continue;
}
}
if(!Pair.empty() && Pair2.empty()) //对子
{
puts("Yes");
continue;
}
if(!Pair.empty() && !Pair2.empty())
{
if(Pair.top() >= Pair2.top())
{
puts("Yes");
continue;
}
}
if(Single.empty() && !Single2.empty()) //单个牌
{
puts("No");
continue;
}
if(!Single.empty() && Single2.empty())
{
puts("Yes");
continue;
}
if(!Single.empty() && !Single2.empty())
{
if(Single.top() >= Single2.top())
{
puts("Yes");
continue;
}
}
puts("No"); //如果以上都不满足，那么A输了，gg。
}
return 0;
}
HDU 4930 Fighting the Landlords --多Trick，较复杂模拟,布布扣,bubuko.com
时间： 2024-08-21 23:05:15