CCPC 2019秦皇岛 Invoker I题 hdu6739

#include
using namespace std;
const int N=1e6+5;
char dict[10][6][4]= {                      //本题的状态与每个组合技能的顺序有关
	{"QQQ","QQQ","QQQ","QQQ","QQQ","QQQ"},      //直接枚举每个组合技能的顺序
	{"QQW","QWQ","QQW","QWQ","WQQ","WQQ"},
	{"QQE","QEQ","QQE","QEQ","EQQ","EQQ"},
	{"WWW","WWW","WWW","WWW","WWW","WWW"},
	{"QWW","QWW","WQW","WWQ","WQW","WWQ"},
	{"WWE","WEW","WWE","WEW","EWW","EWW"},
	{"EEE","EEE","EEE","EEE","EEE","EEE"},
	{"QEE","QEE","EQE","EEQ","EQE","EEQ"},
	{"WEE","WEE","EWE","EEW","EWE","EEW"},
	{"QWE","QEW","WQE","WEQ","EQW","EWQ"}
};

int dp[N][10];
int cal(int a,int b,int x,int y) { //a表示前一个状态（第a个状态）  a x表示第a个组合技第x种顺序
	//计算前后两种状态组合技需要多按几次按键
	if(dict[a][x][0]==dict[b][y][0] && dict[a][x][1] == dict[b][y][1] && dict[a][x][2]==dict[b][y][2])
		return 0;
	if(dict[a][x][1]==dict[b][y][0] && dict[a][x][2] == dict[b][y][1])
		return 1;
	if(dict[a][x][2]== dict[b][y][0])
		return 2;
	return 3;
}
mapmp;
int st[N];
int main() {
	string s;
	mp['Y']=0;
	mp['V']=1;
	mp['G']=2;
	mp['C']=3;
	mp['X']=4;
	mp['Z']=5;
	mp['T']=6;
	mp['F']=7;
	mp['D']=8;
	mp['B']=9;
	while(cin>>s) {
		int ans=2147483647;
		int cc=0;

		for(int i=0; i=1&&s[i]==s[i-1])
				continue;
			else {
				//	t+=s[i];
				st[cc]=mp[s[i]];
				cc++;
			}
		}
//	cout<

参考了这位大佬的博客：https://blog.csdn.net/qq_43461168/article/details/101210999