LIS

耍了这么久，我又回来了。不知道赶不赶得上

题目传送门

C题不会

LIS多种解法

DP法（n*n）

lower/upper_bound优化（n*logn）

方法精髓 int k = lower_bound( t + 1, t + 1 + b, c[ i ] ) - t ;

A

优化后的模板（lower_bound）

int main() {
	int a, b;
	cin >> a;
	while (a--) {
		cin >> b;
		for (int i = 1; i <= b; i++) {
			cin >> c[i];
			t[i] = 999999999;
		}
		int ans = 1;
		for (int i = 1; i <= b; i++) {
			int k = lower_bound(t + 1, t + 1 + b, c[i])-t;
			ans = max(k, ans);
			t[k] = c[i];//更改
		}
		cout << ans << endl;
	}
}

B

本质还是LIS

int main() {
	int a;
	int b;
	int d;
	int t = 0;
	while (scanf("%d", &a)!=EOF) {
		t++;
		int ans = 1;
		for (int i = 1; i <= a; i++) {
			scanf("%d%d", &b, &d);
			c[b] = d;
			dp[i] = 999999999;
		}
		for (int i = 1; i <= a; i++) {
			int k = upper_bound(dp + 1, dp + 1 + a, c[i]) - dp;
			ans = max(k, ans);
			dp[k] = c[i];
		}
		printf("Case %d:\n", t);
		if (ans == 1)
			printf("My king, at most %d road can be built.\n\n", ans);
		else
			printf("My king, at most %d roads can be built.\n\n", ans);
			//注意输出格式加不加s
	}
}

D

模板题，直接写，dp法（n*n）

while (cin >> a&&a) {
		for (int i = 1; i <= a; i++)cin >> c[i];
		for (int i = 1; i <= a; i++)dp[i] = c[i];
		int ans=-999999999;
		for (int i = 1; i <= a; i++) {
			for (int j = 1; j < i; j++) {
				if (c[j] < c[i]) 
					dp[i] = max(dp[i], dp[j] + c[i]);
			}
			ans = max(ans, dp[i]);
		}
		cout << ans << endl;
	}

E

开个数组记录答案就行了

int main() {
	int a, b;
	cin >> a;
	while (a--) {
		cin >> b;
		for (int i = 1; i <= b; i++) {
			cin >> c[i];
			t[i] = 999999999;
		}
		for (int i = 1; i <= b; i++) {
			int k = lower_bound(t + 1, t + 1 + b, c[i])-t;
			p[i] = k;
			t[k] = c[i];//更改
		}
		for (int i = 1; i <= b; i++) {
			if (i == 1)cout << p[i];
			else
				cout << " " << p[i];
		}
		cout << endl;
	}
}

F

根据题意，序列最长上升，下降长度要大于等于b-1；

int main() {
	int a;
	//cin >> a;
	scanf("%d", &a);
	while (a--) {
		int b;
		int ans = 1;
		int ans2 = 1;
		//cin >> b;
		scanf("%d", &b);
		for (int i = 1; i <= b; i++) {
			//cin >> c[i];
			scanf("%d", &c[i]);
			t[i] = 999999999;
			t2[i] = 999999999;
		}
		//最长上升子序列
		for (int i = 1; i <= b; i++) {
			int k = upper_bound(t + 1, t + b + 1, c[i])-t;
			ans = max(ans, k);
			t[k] = c[i];
		}
		reverse(c + 1, c + b + 1);
		//最长下降子序列
		for (int i = 1; i <= b; i++) {
			int k = upper_bound(t2 + 1, t2 + b + 1, c[i]) - t2;
			ans2 = max(ans2, k);
			t2[k] = c[i];
		}
		//cout << ans << " " << ans2 << "?????"<<endl;
		if (ans >= b - 1 || ans2 >= b - 1)printf("YES\n");
		else printf("NO\n");
	}
}

G

思路大魔王，我丢
由题意可以知晓 a [ i ] - a[ j ] >= i - j
移相一下就可以知道

int main() {
	int a;
	int b;
	scanf("%d", &a);
	int p = 1;
	while (a--) {
		printf("Case #%d:\n", p++);
		scanf("%d", &b);
		for (int i = 1; i <= b; i++) {
			scanf("%d", &c[i]);
			c[i] -= i;//相当于移相操作
			dp[i] = 999999999;
		}
		int ans = 0;//注意初始化
		for (int i = 1; i <= b; i++) {
			int k = upper_bound(dp + 1, dp + 1 + b, c[i])-dp;
			ans = max(ans, k);
			dp[k] = c[i];
		}
		printf("%d\n", b - ans);
	}
}