Codeforces Beta Round #87 (Div. 2 Only)

A：水题。。。模拟取最大。。。。

B：由于保证了每个羊最多对应一只狼，所以直接扫描一遍判定就可以了，注意同一个狼别加两次= =。

C：求所有树的最高树高。证明：每个树任意两层的节点肯定不能位于同一个集合，这样得出至少需要最高树高个集合，然后我们把所有的树画在一条线上，每一层对应一个集合，这样可以构造一个包含最高树高个集合的可行解，所以最少需要的集合数为最高树高。

D：裸贪心。。注意那种没有weed的行的处理，开始写晕了，还在pretest 8上WA了一次，无奈之下果断重写了一次好了= =。

A：

#include <iostream>
#include <cstdio>
#include <ctime>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
using namespace std;

#define MP(x, y) make_pair(x, y)
#define FOR(i, x, y) for((i) = (x); (i) <= (y); (i)++)

typedef long long llg;

int n, c, a, b;

int main()
{
    //freopen("data.txt", "r", stdin);
    c = 0;
    int ans = 0;
    scanf("%d", &n);
    while(n--)
    {
        scanf("%d%d", &a, &b);
        c -= a;
        c += b;
        ans = max(ans, c);
    }
    printf("%d\n", ans);
    return  0;
}

　　

B：

#include <iostream>
#include <cstdio>
#include <ctime>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;

#define MP(x, y) make_pair(x, y)
#define FOR(i, x, y) for((i) = (x); (i) <= (y); (i)++)

typedef long long llg;

const int px[4] = {0, 0, 1, -1};
const int py[4] = {1, -1, 0, 0};

int n, m, ans;
char map[15][15];
bool visit[15][15];

int main()
{
    int i, j, k, tx, ty;
    freopen("data.txt", "r", stdin);
    scanf("%d%d", &n, &m);
    for(i = 0; i < n; i++)  scanf(" %s", map[i]);
    ans = 0;
    memset(visit, false, sizeof(visit));
    for(i = 0; i < n; i++)
        for(j = 0; j < m; j++)
        {
            if(map[i][j] != 'P')  continue;
            for(k = 0; k < 4; k++)
            {
                tx = i + px[k];
                ty = j + py[k];
                if(tx<0 || tx>=n || ty<0 || ty>=m)  continue;
                if(map[tx][ty]=='W' && !visit[tx][ty])
                {
                    ans++;
                    visit[tx][ty] = true;
                }
            }
        }
    printf("%d\n", ans);
    return  0;
}

C：

#include <iostream>
#include <cstdio>
#include <ctime>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
using namespace std;

#define MP(x, y) make_pair(x, y)
#define FOR(i, x, y) for((i) = (x); (i) <= (y); (i)++)

typedef long long llg;

const int N = 2010;

vector < vector <int> > g(N);

int n, f[N];

int getLen(int x)
{
    int ct = 1;
    while(f[x] != -1)
    {
        ct++;
        x = f[x];
    }
    return  ct;
}

int main()
{
    int i, ans = 0;
    freopen("data.txt", "r", stdin);
    scanf("%d", &n);
    for(i = 1; i <= n; i++)  scanf("%d", f+i);
    for(i = 1; i <= n; i++)  ans = max(ans, getLen(i));
    printf("%d\n", ans);
    return  0;
}

D:

#include <iostream>
#include <cstdio>
#include <ctime>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;

#define MP(x, y) make_pair(x, y)
#define FOR(i, x, y) for((i) = (x); (i) <= (y); (i)++)

typedef long long llg;

const int N = 160;

int n, m, l[N], r[N], flag[N];
char map[N][N];

int abs(int x)
{
    if(x < 0)  x = -x;
    return  x;
}

int solve()
{
    int i, ans = 0;
    if(l[1] == -1)  flag[1] = 0;
    else  flag[1] = r[1];
    for(i = n; i > 0; i--)
        if(r[i] != -1)
            break;
    n = i;
    if(n == 0)  return  0;
    if(r[1] != -1)  ans += r[1];
    for(i = 2; i <= n; i++)
    {
        ans++;
        if(i%2 == 0)
        {
            if(r[i] != -1)
                ans += abs(r[i]-flag[i-1]);
            flag[i] = l[i];
        }
        else
        {
            if(l[i] != -1)
                ans += abs(l[i]-flag[i-1]);
            flag[i] = r[i];
        }
        if(flag[i] == -1)  flag[i] = flag[i-1];
        ans += (r[i]-l[i]);
    }
    return  ans;
}

int main()
{
    int i, j, ans;
    freopen("data.txt", "r", stdin);
    scanf("%d%d", &n, &m);
    for(i = 1; i <= n; i++)
        scanf(" %s", map[i]);
    memset(l, -1, sizeof(l));
    memset(r, -1, sizeof(r));
    for(i = 1; i <= n; i++)
    {
        for(j = 0; j < m; j++)
            if(map[i][j] == 'W')
            {
                l[i] = j;
                break;
            }
        for(j = m-1; j >= 0; j--)
            if(map[i][j] == 'W')
            {
                r[i] = j;
                break;
            }
    }
    ans = solve();
    printf("%d\n", ans);
    return  0;
}