2020 计蒜之道 预赛 第一场 A、B

2020计蒜之道 预赛第一场 A、B

链接：https://www.jisuanke.com/contest/11344/challenges

A. 五子棋

枚举每个空格点的位置，然后判水平垂直以及两条对角线即可。判的时候可以沿那个方向往两边走，有连续棋子多于4个的话就获胜。

#include <bits/stdc++.h>
using namespace std;
int n = 24; 
char mmap[30][30];
char go;
struct Point
{
	int x, y;
};
vector<Point> v;
bool tran(int x, int y)
{
	int cnt = 0, nx = x, ny = y - 1;
	while(ny >= 0 && mmap[nx][ny] == go)
	{
		cnt++;
		ny--;
	}
	ny = y + 1;
	while(ny <= n && mmap[nx][ny] == go)
	{
		cnt++;
		ny++;
	}
	if(cnt >= 4) return 1;
	else return 0;
}
bool vertical(int x, int y)
{
	int cnt = 0, nx = x - 1, ny = y;
	while(nx >= 0 && mmap[nx][ny] == go)
	{
		cnt++;
		nx--;
	}
	nx = x + 1;
	while(nx <= n && mmap[nx][ny] == go)
	{
		cnt++;
		nx++;
	}
	if(cnt >= 4) return 1;
	else return 0;
}
bool MainDiagonal(int x, int y)
{
	int cnt = 0, nx = x - 1, ny = y - 1;
	while(nx >= 0 && ny >= 0 && mmap[nx][ny] == go)
	{
		cnt++;
		nx--;
		ny--;
	}
	nx = x + 1, ny = y + 1;
	while(nx <= n && ny <= n && mmap[nx][ny] == go)
	{
		cnt++;
		nx++;
		ny++;
	}
	if(cnt >= 4) return 1;
	else return 0;
}
bool SubDiagonal(int x, int y)
{
	int cnt = 0, nx = x - 1, ny = y + 1;
	while(nx >= 0 && ny <= n && mmap[nx][ny] == go)
	{
		cnt++;
		nx--;
		ny++;
	}
	nx = x + 1, ny = y - 1;
	while(nx <= n && ny >= 0 && mmap[nx][ny] == go)
	{
		cnt++;
		nx++;
		ny--;
	}
	cout << cnt << endl;
	if(cnt >= 4) return 1;
	else return 0;
}
int main()
{
	for(int i = 0; i < 25; i++)
	{
		scanf("%s", mmap[i]);
	}
	int w = 0, b = 0;
	for(int i = 0; i < 25; i++)
	{
		for(int j = 0; j < 25; j++)
		{
			if(mmap[i][j] == 'o') w++;
			else if(mmap[i][j] == 'x') b++; 
		}
	}

	if(w == b)//轮到黑棋走 
	{
		go = 'x';
	}
	else go = 'o';
	for(int i = 0; i < 25; i++)
	{
		for(int j = 0; j < 25; j++)
		{
			if(mmap[i][j] == '.')
			{
				if(tran(i, j) || vertical(i, j) || MainDiagonal(i, j) || SubDiagonal(i, j))
				{
					v.push_back({i, j});
				}
			}
		}
	}
	if(!v.size()) cout << "tie" << endl;
	else 
		for(int i = 0; i < v.size(); i++) cout << v[i].x << ' ' << v[i].y << endl;
	return 0;
}

B. 染色（简单）

没在区间的直接染w和b较大的那个颜色，对于每个区间，由于彼此互不相交，直接比较“染同一个颜色然后获得bouns”以及“每个元素染w和b较大那个”然后取最大值累加到答案即可。

#include <bits/stdc++.h>
#define int long long//!!!!!
using namespace std;
int n, m, w[300005], b[300005], t[300005], l[300005], r[300005], c[300005];
bool seg[300005];
long long ans = 0; 
signed main()
{
	memset(seg, 0, sizeof(seg));
	cin >> n >> m;
	for(int i = 1; i <= n; i++) cin >> b[i];
	for(int i = 1; i <= n; i++) cin >> w[i];
	for(int i = 1; i <= m; i++)
	{
		cin >> t[i] >> l[i] >> r[i] >> c[i];
		for(int j = l[i]; j <= r[i]; j++) seg[j] = 1;//在区间 
	}
	for(int i = 1; i <= n; i++)
	{
		if(!seg[i]) ans += max(b[i], w[i]);
	}
	for(int i = 1; i <= m; i++)
	{
		int mmax = 0, black = 0, white = 0;
		for(int j = l[i]; j <= r[i]; j++)
		{
			black += b[j];
			white += w[j];
			mmax += max(b[j], w[j]);
		}
		if(t[i] == 1)
		{
			ans += max(c[i] + black, mmax);
		}
		else
		{
			ans += max(c[i] + white, mmax);
		}
	}
	cout << ans << endl;
}