「NOIP模拟赛」Round 2

Tag

递推,状压DP,最短路

A. 篮球比赛1

题面

\(Milky\ Way\)的代码

#include <cstdio>

const int N = 2000, xzy = 1e9 + 7;
int a[N], f[N][N], g[N][N], n, m = 1023, ans;

//f[i][j]表示前i个数的xor和为j的方案数
//g[i][j]表示后n-i+1个数的and和为j的方案数

int main() {
	freopen("basketball1.in", "r", stdin);
	freopen("basketball1.out", "w", stdout);

	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
	f[0][0] = 1;
	for (int i = 1; i <= n; ++i) { //前i个数
		for (int j = 0; j <= m; ++j)
			(f[i][j] += f[i-1][j]) %= xzy;
		for (int j = 0; j <= m; ++j) //前i-1个数的xor和为j
			(f[i][j^a[i]] += f[i-1][j]) %= xzy;
	}
	for (int i = n; i >= 1; --i) { //后n-i+1个数
		g[i][a[i]] = 1;
		for (int j = 0; j <= m; ++j)
			(g[i][j] += g[i+1][j]) %= xzy;
		for (int j = 0; j <= m; ++j) //后n-i个数的and和为j
			(g[i][j&a[i]] += g[i+1][j]) %= xzy;
	}
	for (int i = 1; i < n; ++i) //前i个数
		for (int j = 0; j <= m; ++j) //前i-1个数的xor和
			ans = (ans + 1LL * f[i-1][j] * g[i+1][j^a[i]] % xzy) % xzy;
	printf("%d\n", ans);

	fclose(stdin);
	fclose(stdout);
	return 0;
}

B. 篮球比赛2

题面

\(Milky\ Way\)的代码

#include <cstdio>

const int N = 25, M = 1048580, XZY = 1e9 + 7;
int f[N][M], n, k, l, m, ans;

int main() {
	freopen("basketball2.in", "r", stdin);
	freopen("basketball2.out", "w", stdout);

	scanf("%d%d%d", &n, &k, &l);
	m = (1 << k) - 1;
	f[0][0] = 1;
	for (int i = 1; i <= n; ++i) { //前i个数
		if (l > k) for (int j = 0; j <= m; ++j) //前i-1个数能取到的状态
			f[i][j] = 1LL * f[i-1][j] * (l - k) % XZY; //x>k
		for (int j = 0; j <= m; ++j) {
			(f[i][j] += f[i-1][j]) %= XZY; //x=0
			if (f[i-1][j]) for (int x = 1; x <= k && x <= l; ++x) //注意x要同时小于k和l
				(f[i][(j|(j<<x)|(1<<x-1))&m] += f[i-1][j]) %= XZY; //1<=x<=k
		}
	}
	for (int i = 0; i <= m; ++i)
		if (i >> k - 1 & 1) (ans += f[n][i]) %= XZY;
	printf("%d\n", ans);

	fclose(stdin);
	fclose(stdout);
	return 0;
}

C. 密室逃脱

题面

\(Lmsh7\)的代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
using std::sort;
using std::queue;
using std::max;
using std::min;

typedef long long LL;

const int N = 105;
const int dx[] = {1, -1, 0, 0};
const int INF = 0X3f3f3f3f;
const int dy[] = {0, 0, -1, 1};

int n, m, cnt, stx, sty, edx, edy, ans = INF;
int a[N][N], f[N][N][11];
char s[N];

struct Node {
	int x, y, m;
} b[N * N];

inline void bfs() {
	memset(f, -1, sizeof(f));
	queue <Node> q;
	q.push(Node{stx, sty, 0});
	f[stx][sty][0] = 0;
	while(!q.empty()) {
		int ux = q.front().x, uy = q.front().y, um = q.front().m;
//		printf("%d %d %d\n", ux, uy, um);
		q.pop();
		for(int i = 0; i < 4; ++i) {
			int nowx = ux + dx[i], nowy = uy + dy[i], nowm = um;
			if(nowx < 1 || nowx > n || nowy < 1 || nowy > n) continue;
			if(a[nowx][nowy] == -1 || f[nowx][nowy][nowm] != -1) continue;
			if(a[nowx][nowy] == nowm + 1) ++nowm;
			f[nowx][nowy][nowm] = f[ux][uy][um] + 1;
			q.push(Node{nowx, nowy, nowm});
		}
	}
	return;
}

void dfs(int x, int y) {//x -> cnt, y -> sum
	if(x == cnt + 1) {
		bfs();
//		printf("f[%d][%d][%d] = %d\n", edx, edy, m, f[edx][edy][m]);
		if(f[edx][edy][m] != -1) {
			ans = min(ans, f[edx][edy][m] + y);
		}

		return;
	}
	a[b[x].x][b[x].y] = -1;
	dfs(x + 1, y);
	a[b[x].x][b[x].y] = 0;
	dfs(x + 1, y + 1);
	return;
}

int main() {
	freopen("maze.in", "r", stdin);
	freopen("maze.out", "w", stdout);

	scanf("%d %d", &n, &m);
	for(int i = 1; i <= n; ++i) {
		scanf("%s", s + 1);
		for(int j = 1; j <= n; ++j) {
			if(s[j] == 'K') {
				stx = i, sty = j;//璧风偣
			} else if(s[j] == 'T') {
				edx = i, edy = j;//缁堢偣
			} else if(s[j] == 'S') {
				b[++cnt].x = i;
				b[cnt].y = j;//鐗规畩鎴块棿
			} else if(s[j] == '.') {
				a[i][j] = 0;
			} else if(s[j] == '#') {
				a[i][j] = -1;
			} else {
				a[i][j] = s[j] - '0';
			}
		}
	}
	dfs(1, 0);
	if(ans == INF) puts("impossible");
	else printf("%d\n", ans);

	fclose(stdin);
	fclose(stdout);
	return 0;
}