洛谷 P1335 / UOJ #123 - [NOI2013] 小Q的修炼（提交答案题）

洛谷题面传送门 & UOJ 题面传送门

一道相对来说难度较低的提答题。

首先碰到提答题我们肯定不能思考一般性解法，只能就事论事，对着对应的数据设计出相应的程序。纵观 10 组数据，我们可以获得一些的结论：

  • 对于所有跳转事件，那些目标位置 \(<\) 当前位置的事件，都形如 i c 0 c 1 x 0，也就是如果 \(0<1\) 跳转到 \(x\)，否则跳转到 \(0\)​，显然这种事件只能选择前者，也就是说本题的结构实际上是一个有限状态自动机，不会成环。
  • 在第 \(2,3,7,8,9,10\) 中都出现了一个类似的结构，也就是一段 v 3 x1, v 4 x2, ..., v 12 x10 后跟一堆将变量 \(v_3,v_4,\cdots,v_{12}\) 累加到 \(v_1\) 上的操作，不难发现这可以视作，有若干个大礼包，每个大礼包可以选择或者不选择，如果选择则需要将 \(v_3,v_4,\cdots,v_{12}\) 都加上一个固定的值，隔一段时间以后令答案加上 \(\sum\limits_{i=3}^{12}|v_i|\)，我们称之为 A 结构。
  • 在第 \(4,5,6,7,8,9,10\)​ 中也出现了一个类似的结构，也就是所有与 \(v_2\)​ 有关的操作，稍加分析可以发现这是一个背包模型，变量 \(2\)​ 的值就是剩余体积，即，有若干个大礼包，每个大礼包有两个三个属性 \(to_i,v_i,w_i\)​，每进入一个礼包，如果剩余体积 \(<v_i\)​ 则直接转到 \(to_i\)，否则你可以选择到 \(i+1\)，并花费 \(w_i\) 的体积或者 \(v_i\) 的价值，也可以选择到 \(to_i\)。我们称之为 B 结构。

做出这些小观察后本题就容易许多了。

对于第一组数据，选择操作很少，直接爆搜即可。这里直接给出答案。

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对于第二组数据，是一个有 25 个大礼包的 A 结构，同样直接 \(2^{25}\) 暴力即可。

const int MAXN = 3.5e5;
int n, m;
int val[15];
struct number {
	int opt, id;
	void read() {
		static char str[5]; scanf("%s%d", str + 1, &id);
		opt = (str[1] == 'v');
	}
	int operator () () {return (opt) ? val[id] : id;}
};
struct qry {int opt, A, B; number C, D;} a[MAXN + 5];
int main() {
	freopen("train2.in", "r", stdin);
	freopen("train2.out", "w", stdout);
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		static char str[5]; scanf("%s", str + 1);
		if (str[1] == 'v') {
			static char op[5]; scanf("%d%s", &a[i].A, op + 1);
			a[i].B = (op[1] == '-'); // 0 for + and 1 for -
			a[i].C.read(); a[i].opt = 1;
		} else if (str[1] == 's') {
			scanf("%d%d", &a[i].A, &a[i].B);
			a[i].opt = 2;
		} else {
			a[i].C.read(); a[i].D.read();
			scanf("%d%d", &a[i].A, &a[i].B);
			a[i].opt = 3;
		}
	}
	int sum = 0;
	for (int i = 1; i <= n; i++) sum += (a[i].opt == 2);
	int mx = 0, msk = -1;
	for (int i = 0; i < (1 << sum); i++) {
		fill0(val); int cur = 0;
		for (int j = 1; j <= n; j++) {
			if (a[j].opt == 1) {
				if (a[j].B) val[a[j].A] -= a[j].C();
				else val[a[j].A] += a[j].C();
			} else if (a[j].opt == 2) {
				if (~i >> cur & 1) j = a[j].A - 1;
				else j = a[j].B - 1;
				cur++;
			} else {
				if (a[j].C() < a[j].D()) j = a[j].A - 1;
				else j = a[j].B - 1;
			}
		}
		if (i % 1000000 == 0) cerr << i << " " << val[1] << endl;
		if (val[1] > mx) mx = val[1], msk = i;
	}
	cerr << msk << " " << mx << endl;
	for (int i = 0; i < 25; i++) printf("%d\n", (msk >> i & 1) + 1);
	return 0;
}

对于第三组数据，相当于几百个 A 类结构，每个 A 类结构里有大概 \(10\) 个大礼包，对每个 A 类结构暴力求解加起来即可。

const int MAXN = 3.5e5;
int n, m, pos[MAXN + 5][15];
int val[15];
struct number {
	int opt, id;
	void read() {
		static char str[5]; scanf("%s%d", str + 1, &id);
		opt = (str[1] == 'v');
	}
	int operator () () {return (opt) ? val[id] : id;}
};
struct qry {int opt, A, B; number C, D;} a[MAXN + 5];
int main() {
	freopen("train3.in", "r", stdin);
	freopen("train3.out", "w", stdout);
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		static char str[5]; scanf("%s", str + 1);
		if (str[1] == 'v') {
			static char op[5]; scanf("%d%s", &a[i].A, op + 1);
			a[i].B = (op[1] == '-'); // 0 for + and 1 for -
			a[i].C.read(); a[i].opt = 1;
		} else if (str[1] == 's') {
			scanf("%d%d", &a[i].A, &a[i].B);
			a[i].opt = 2;
		} else {
			a[i].C.read(); a[i].D.read();
			scanf("%d%d", &a[i].A, &a[i].B);
			a[i].opt = 3;
		}
	}
	vector<vector<int> > vec;
	for (int i = 1; i <= n; ) {
		if (a[i].opt == 2) {
			vector<int> tmp;
			for (int j = a[i].A; j < a[i].B; j++) tmp.pb(a[j].C());
			vec.pb(tmp); i = a[i].B;
		} else {
			ll mx = -1; int msk = -1;
			for (int j = 0; j < (1 << vec.size()); j++) {
				static ll ss[15]; fill0(ss);
				for (int k = 0; k < vec.size(); k++) if (j >> k & 1) {
					for (int l = 0; l < vec[k].size(); l++)
						ss[l] += vec[k][l];
				}
				ll sum = 0;
				for (int k = 0; k < 10; k++) sum += abs(ss[k]);
				if (sum > mx) mx = sum, msk = j;
			}
//			printf("! %lld %d %d\n", mx, msk, (int)vec.size());
			for (int j = 0; j < vec.size(); j++) {
				if (msk >> j & 1) puts("1");
				else puts("2");
			}
			vec.clear();
			i += 50;
		}
	}
	return 0;
}

对于第四、五、六组数据，相当于一个 B 类结构，从后往左背包即可。

const int MAXN = 3.5e5;
int n, m, val[15];
struct number {
	int opt, id;
	void read() {
		static char str[5]; scanf("%s%d", str + 1, &id);
		opt = (str[1] == 'v');
	}
	int operator () () {return (opt) ? val[id] : id;}
};
struct qry {int opt, A, B; number C, D;} a[MAXN + 5];
int id[MAXN + 5], idcnt = 0, bel[MAXN + 5], nxt[MAXN + 5];
ll dp[2005][10005]; int to[2005][10005], cst[MAXN + 5], v[MAXN + 5];
vector<int> vec;
void work(int cur, int lft) {
	if (cur > idcnt) return;
	if (to[cur][lft]) vec.pb(1), work(cur + 1, lft - cst[cur + 1]);
	else {
		if (lft >= cst[cur + 1]) vec.pb(2);
		work(nxt[cur], lft);
	}
}
int main() {
	freopen("train6.in", "r", stdin);
	freopen("train6.out", "w", stdout);
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		static char str[5]; scanf("%s", str + 1);
		if (str[1] == 'v') {
			static char op[5]; scanf("%d%s", &a[i].A, op + 1);
			a[i].B = (op[1] == '-'); // 0 for + and 1 for -
			a[i].C.read(); a[i].opt = 1;
		} else if (str[1] == 's') {
			scanf("%d%d", &a[i].A, &a[i].B);
			a[i].opt = 2;
		} else {
			a[i].C.read(); a[i].D.read();
			scanf("%d%d", &a[i].A, &a[i].B);
			a[i].opt = 3;
		}
	}
	for (int i = 1; i <= n; i++) if (a[i].opt == 2) id[i] = ++idcnt;
	bel[n + 1] = idcnt + 1;
//	cerr << idcnt << endl;
	for (int i = n; i; i--) bel[i] = ((a[i].opt == 2) ? id[i] : bel[i + 1]);
	for (int i = 1; i <= n; i++) if (a[i].opt == 2) nxt[id[i]] = bel[a[i].B];
	for (int i = 4; i <= n; i++) if (a[i].opt == 1) {
		if (a[i].A == 2) cst[bel[i]] = a[i].C();
		else v[bel[i]] = a[i].C();
	}
	for (int i = idcnt; i; i--) for (int j = 0; j <= 10000; j++) {
		if (j < cst[i + 1]) dp[i][j] = dp[nxt[i]][j];
		else {
			if (dp[nxt[i]][j] < dp[i + 1][j - cst[i + 1]] + v[i + 1]) {
				dp[i][j] = dp[i + 1][j - cst[i + 1]] + v[i + 1];
				to[i][j] = 1;
			} else dp[i][j] = dp[nxt[i]][j];
		}
	}
	ll mx = 0; int mxp = -1;
	for (int i = 0; i <= 10000; i++) if (mx < dp[1][i]) mx = dp[1][i], mxp = i;
	work(1, mxp); cerr << mx << endl;
	for (int x : vec) printf("%d\n", x);
	return 0;
}

对于第七、八、九、十组数据，相当于一个 B 类结构，但 B 类结构每个礼包内部又是一个 A 类结构，同样道理，先对每个 A 类结构跑一遍第三组数据的求解方法找出其代价，然后再跑四、五、六即可。

const int MAXN = 3.6e5;
int n, m, val[15];
struct number {
	int opt, id;
	void read() {
		static char str[5]; scanf("%s%d", str + 1, &id);
		opt = (str[1] == 'v');
	}
	int operator () () {return (opt) ? val[id] : id;}
};
struct qry {int opt, A, B; number C, D;} a[MAXN + 5];
int id[MAXN + 5], idcnt = 0, bel[MAXN + 5], nxt[MAXN + 5];
ll dp[2005][10005]; int to[2005][10005];
ll cst[MAXN + 5], v[MAXN + 5];
vector<int> vec;
bool flg[MAXN + 5];
vector<vector<int> > tt[MAXN + 5];
int mskv[MAXN + 5];
void deal(int id) {
	for (int i = 0; i < tt[id].size(); i++)
		vec.pb((mskv[id] >> i & 1) ? 1 : 2);
}
void work(int cur, int lft) {
	if (cur > idcnt) return;
	if (to[cur][lft]) vec.pb(1), deal(cur + 1), work(cur + 1, lft - cst[cur + 1]);
	else {
		if (lft >= cst[cur + 1]) vec.pb(2);
		work(nxt[cur], lft);
	}
}
int main() {
	freopen("train10.in", "r", stdin);
	freopen("train10.out", "w", stdout);
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		static char str[5]; scanf("%s", str + 1);
		if (str[1] == 'v') {
			static char op[5]; scanf("%d%s", &a[i].A, op + 1);
			a[i].B = (op[1] == '-'); // 0 for + and 1 for -
			a[i].C.read(); a[i].opt = 1;
		} else if (str[1] == 's') {
			scanf("%d%d", &a[i].A, &a[i].B);
			a[i].opt = 2;
		} else {
			a[i].C.read(); a[i].D.read();
			scanf("%d%d", &a[i].A, &a[i].B);
			a[i].opt = 3;
		}
	}
	for (int i = 1; i <= n; i++) if (a[i].opt == 2) {
		flg[i] = 1;
		for (int j = 1; j <= 10; j++) flg[i] &= (a[i + j].opt == 1);
	} 
	for (int i = 1; i <= n; i++) if (a[i].opt == 2 && !flg[i]) id[i] = ++idcnt;
	bel[n + 1] = idcnt + 1;
//	for (int i = 1; i <= n; i++) cerr << id[i] << endl;
//	cerr << idcnt << endl;
	for (int i = n; i; i--) bel[i] = ((a[i].opt == 2 && !flg[i]) ? id[i] : bel[i + 1]);
	for (int i = 1; i <= n; i++) if (a[i].opt == 2 && !flg[i]) nxt[id[i]] = bel[a[i].B];
	for (int i = 1; i <= n; i++) if (a[i].opt == 2 && flg[i]) {
		vector<int> tmp;
		for (int j = 1; j <= 10; j++) tmp.pb(a[i + j].C());
		tt[bel[i]].pb(tmp);
	}
	for (int i = 4; i <= n; i++) if (a[i].opt == 1 && a[i].A == 2)
		cst[bel[i]] = a[i].C();
	for (int i = 1; i <= idcnt + 1; i++) {
		ll mx = -1; int msk = -1;
		for (int j = 0; j < (1 << tt[i].size()); j++) {
			static ll vals[15]; fill0(vals);
			for (int k = 0; k < tt[i].size(); k++) if (j >> k & 1)
				for (int l = 0; l < 10; l++) vals[l] += tt[i][k][l];
			ll ss = 0;
			for (int l = 0; l < 10; l++) ss += abs(vals[l]);
			if (ss > mx) mx = ss, msk = j;
		}
		v[i] = mx; mskv[i] = msk;
	}
	for (int i = 1; i <= idcnt; i++) cerr << v[i] << endl;
	for (int i = idcnt; i; i--) for (int j = 0; j <= 1000; j++) {
		if (j < cst[i + 1]) dp[i][j] = dp[nxt[i]][j];
		else {
			if (dp[nxt[i]][j] < dp[i + 1][j - cst[i + 1]] + v[i + 1]) {
				dp[i][j] = dp[i + 1][j - cst[i + 1]] + v[i + 1];
				to[i][j] = 1;
			} else dp[i][j] = dp[nxt[i]][j];
		}
	}
	ll mx = 0; int mxp = -1;
	for (int i = 0; i <= 1000; i++) if (mx < dp[1][i]) mx = dp[1][i], mxp = i;
	work(1, mxp); cerr << mx << endl;
	for (int x : vec) printf("%d\n", x);
	return 0;
}
posted @ 2022-02-07 23:23  tzc_wk  阅读(715)  评论(0)    收藏  举报