Codeforces 2017 JUST Programming Contest 3.0 D. Dice Game

Dice Game

• 骰子初始状态顶点为1，每次骰子只可以转90度，即顶点的对面无法到达(骰子的每个面与其对面数字和为7)
• 题中要求最小步数，初始化数组各元素为0x3f3f3f3f

dp[score][top]//得分为score，顶点为top时最小步数
dp[0][1] = 0;
//骰子每次只能转一次
//打好表，在六个顶点，score中找最小步骤，输出即可
    for(int i = 0; i <= 10000;i++){
        for(int j = 1; j < 7; j++){
            //f[i][j]翻转,顶点为j 
            for(int k = 1; k < 7; k++){
                if(k != j && k + j != 7) f[i + k][k] = min(f[i + k][k],f[i][j] + 1);
            }
        }
    }	
	cin >> t;
	int l = 0x3f3f3f3f;
	for(int i = 1; i < 7; i++) l = min(f[t][i],l);
	if(l != inf)cout << l << endl;
	else cout << "-1" << endl;	
	
//bfs() 到达各个点的最小步数，可以增加一个vis数组，广度优先搜索即可
    void bfs(){
        node cur,t;
        cur.sum = cur.step = 0;
        cur.top = 1;
        queue<node> q;
        q.push(cur);
        vis[cur.sum] = true;

        while(!q.empty()){
            cur = q.front();
            q.pop();
            if(cur.sum == 10000) return;
            t.step = cur.step + 1;
            for(int i = 1; i < 7; i++){
                //翻转
                if(i != cur.top && i + cur.top != 7){
                    t.top = i;
                    t.sum = cur.sum + i;
                    if(!vis[t.sum]){
                        f[t.sum][t.top] = t.step;
                        vis[t.sum] = true;
                        q.push(t);
                    } 
                } 
            }
        }
    }
	cin >> t;
	if(!vis[t]) cout << "-1" << endl;
	else{
		l = inf;
		for(int i = 1; i < 7; i++) l = min(f[t][i],l);
		cout << l << endl;			
	}