水 A - Alena's Schedule

/************************************************

* Author        :Running_Time

* Created Time  :2015/10/12 星期一 16:49:42

* File Name     :A.cpp

 ************************************************/

#include <cstdio>

#include <algorithm>

#include <iostream>

#include <sstream>

#include <cstring>

#include <cmath>

#include <string>

#include <vector>

#include <queue>

#include <deque>

#include <stack>

#include <list>

#include <map>

#include <set>

#include <bitset>

#include <cstdlib>

#include <ctime>

using namespace std;

#define lson l, mid, rt << 1

#define rson mid + 1, r, rt << 1 | 1

typedef long long ll;

const int N = 1e2 + 10;

const int INF = 0x3f3f3f3f;

const int MOD = 1e9 + 7;

const double EPS = 1e-8;

int a[N];

int main(void)    {

    int n;	scanf ("%d", &n);

	for (int i=1; i<=n; ++i)	{

		scanf ("%d", &a[i]);

	}

	int ans = 0, i = 1;

	bool fir = true;

	while (i <= n)	{

		if (a[i] == 0)	{

			if (fir)	{

				i++;	continue;

			}

			else	{

				int j = i + 1;

				while (j <= n)	{

					if (a[j] == 0)	{

						j++;

					}

					else	break;

				}

				if (j == n + 1)	break;

				if (j - i >= 2)	{

                    i = j;	continue;

				}

				else	{

					ans += j - i;

					i = j;

				}

			}

		}

		else	{		//a[i] = 1;

			fir = false;

            ans++;	i++;

		}

	}

	printf ("%d\n", ans);

    return 0;

}

　　

水 B - Laurenty and Shop

/************************************************

* Author        :Running_Time

* Created Time  :2015/10/12 星期一 16:49:42

* File Name     :B.cpp

 ************************************************/

#include <cstdio>

#include <algorithm>

#include <iostream>

#include <sstream>

#include <cstring>

#include <cmath>

#include <string>

#include <vector>

#include <queue>

#include <deque>

#include <stack>

#include <list>

#include <map>

#include <set>

#include <bitset>

#include <cstdlib>

#include <ctime>

using namespace std;

#define lson l, mid, rt << 1

#define rson mid + 1, r, rt << 1 | 1

typedef long long ll;

const int N = 55;

const int INF = 0x3f3f3f3f;

const int MOD = 1e9 + 7;

const double EPS = 1e-8;

int a1[N], a2[N], b[N];

int sum1[N], sum2[N];

bool vis[N];

int main(void)    {

	int n;	scanf ("%d", &n);

	for (int i=1; i<n; ++i)	{

		scanf ("%d", &a1[i]);

	}

	for (int i=1; i<n; ++i)	{

		scanf ("%d", &a2[i]);

	}

	for (int i=1; i<=n; ++i)	{

		scanf ("%d", &b[i]);

	}

	if (n == 2)	{

		printf ("%d\n", a1[1] + a2[1] + b[1] + b[2]);

		return 0;

	}

	for (int i=1; i<n; ++i)	{

		sum1[i] = sum1[i-1] + a1[i];

	}

	for (int i=n-1; i>=1; --i)	{

		sum2[i] = sum2[i+1] + a2[i];

	}

	int mn = INF, id = 0;

	int ans = 0;

	for (int i=0; i<n; ++i)	{

        if (sum1[i] + b[i+1] + sum2[i+1] < mn)	{

			mn = sum1[i] + b[i+1] + sum2[i+1];

			id = i;

        }

	}

	vis[id] = true;	ans += mn;

	mn = INF, id = 0;

	for (int i=0; i<n; ++i)	{

		if (sum1[i] + b[i+1] + sum2[i+1] < mn && !vis[i])	{

			mn = sum1[i] + b[i+1] + sum2[i+1];

			id = i;

		}

	}

	ans += mn;

	printf ("%d\n", ans);

    return 0;

}

　　

暴力+队列 C - Gennady the Dentist

题意：小孩子排队看医生，有一定的连锁反应，问医生最后能治好多少人

分析：开始想用D保存d[]的和，无奈被hack掉了，还是老老实实用队列吧

/************************************************

* Author        :Running_Time

* Created Time  :2015/10/12 星期一 16:49:42

* File Name     :C.cpp

 ************************************************/

#include <cstdio>

#include <algorithm>

#include <iostream>

#include <sstream>

#include <cstring>

#include <cmath>

#include <string>

#include <vector>

#include <queue>

#include <deque>

#include <stack>

#include <list>

#include <map>

#include <set>

#include <bitset>

#include <cstdlib>

#include <ctime>

using namespace std;

#define lson l, mid, rt << 1

#define rson mid + 1, r, rt << 1 | 1

typedef long long ll;

const int N = 4e3 + 10;

const int INF = 0x3f3f3f3f;

const int MOD = 1e9 + 7;

const double EPS = 1e-8;

int v[N], d[N], p[N];

bool dead[N];

int main(void)    {

	int n;	scanf ("%d", &n);

	for (int i=1; i<=n; ++i)	{

		scanf ("%d%d%d", &v[i], &d[i], &p[i]);

		dead[i] = false;

	}

	vector<int> ans;

	queue<int> Q;

	for (int i=1; i<=n; ++i)	{

		if (!dead[i])	{

			ans.push_back (i);

			for (int j=i+1; j<=n; ++j)	{

				if (dead[j])	continue;

				p[j] -= v[i];	v[i]--;

				if (p[j] < 0)	{

					dead[j] = true;

					Q.push (j);

				}

				if (v[i] <= 0)	break;

			}

		}

		while (!Q.empty ())	{

			int x = Q.front ();	Q.pop ();

			for (int j=x+1; j<=n; ++j)	{

				if (dead[j])	continue;

				p[j] -= d[x];

				if (p[j] < 0)	{

					dead[j] = true;

					Q.push (j);

				}

			}

		}

	}

	printf ("%d\n", ans.size ());

	for (int i=0; i<ans.size (); ++i)	{

		printf ("%d%c", ans[i], i == ans.size () - 1 ? '\n' : ' ');

	}

    return 0;

}

　　

BFS D - Phillip and Trains

题意：一个人从左边走到右边，同时有列车从右边开过来，问这个人能否达到右边没有碰到列车

分析：简答的BFS，车看作不动，人相对车有一个相对运动，还是要用vis标记已访问过的点，比赛时这点忘了，MLE了

/************************************************

* Author        :Running_Time

* Created Time  :2015/10/12 星期一 16:49:42

* File Name     :D.cpp

 ************************************************/

#include <cstdio>

#include <algorithm>

#include <iostream>

#include <sstream>

#include <cstring>

#include <cmath>

#include <string>

#include <vector>

#include <queue>

#include <deque>

#include <stack>

#include <list>

#include <map>

#include <set>

#include <bitset>

#include <cstdlib>

#include <ctime>

using namespace std;

#define lson l, mid, rt << 1

#define rson mid + 1, r, rt << 1 | 1

typedef long long ll;

const int N = 1e2 + 10;

const int INF = 0x3f3f3f3f;

const int MOD = 1e9 + 7;

const double EPS = 1e-8;

char maze[4][N];

bool vis[4][N];

int dx[3] = {-1, 0, 1};

int n, k;

bool BFS(int sx, int sy)	{

	queue<pair<int, int> > Q;	Q.push (make_pair (sx, sy));

	memset (vis, false, sizeof (vis));  vis[sx][sy] = true;

    while (!Q.empty ())	{

		int x = Q.front ().first, y = Q.front ().second;	Q.pop ();

        if (y >= n)	return true;

        if (y + 1 <= n && maze[x][y+1] != '.')	continue;

        for (int i=0; i<3; ++i)	{

			int tx = x + dx[i], ty = y + 1;

			if (tx < 1 || tx > 3 || maze[tx][ty] != '.')	continue;

            ty += 1;

            if (ty <= n && maze[tx][ty] != '.')	continue;

            ty += 1;

            if (ty <= n && maze[tx][ty] != '.')	continue;

            if (vis[tx][ty])    continue;

            vis[tx][ty] = true;

            Q.push (make_pair (tx, ty));

        }

	}

	return false;

}

int main(void)    {

	int T;	scanf ("%d", &T);

	while (T--)	{

		scanf ("%d%d", &n, &k);

		for (int i=1; i<=3; ++i)	{

			scanf ("%s", maze[i] + 1);

		}

		int sx = 0, sy = 0;

		for (int i=1; i<=3; ++i)	{

            if (maze[i][1] == 's')	{

				sx = i;	sy = 1;

				break;

            }

		}

        for (int i=n+1; i<=n+6; ++i)    {

            maze[1][i] = maze[2][i] = maze[3][i] = '.';

        }

		printf ("%s\n", BFS (sx, sy) ? "YES" : "NO");

	}

    return 0;

}

　　

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