SRM 578 DIV 2

250:

简单题目：

500：

题意：

给定一个矩形，里面要么是"v"表示，要么是"."，v表示可能是g,也可能是d，如果是g的话，那么它的哈弗曼距离dis之内如果是v的话，一定是g。求有多少种满足条件的可能数。

思路：

将每一个块分出来，自这一联通块里面，所有的v要么是g，要么d,bfs把所有的快求出来，假设为n，则最后的总数为2^n - 1

View Code

#line 5 "GooseInZooDivTwo.cpp"
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll long long
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);
#define tr(container, it) for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)

#define M 107
#define N 107
using namespace std;

const ll mod = 1000000007;

struct node
{
    int x,y;
}nd;
int n,m;

vector<string> tmp;

int Abs(int x)
{
    if (x >= 0) return x;
    else return (-x);
}
void bfs(int x,int y,int k)
{
    queue<node>q;
    nd.x = x; nd.y = y;
    q.push(nd);
    int i,j;
    while (!q.empty())
    {
        node u = q.front(); q.pop();

        for (i = max(0,u.x - k); i <= min(n - 1, u.x + k); ++i)
        {
            for (j = max(0,u.y - k); j <= min(m - 1,u.y + k); ++j)
            {
                if (Abs(u.x - i) + Abs(u.y - j) <= k && tmp[i][j] == 'v')
                {
                    tmp[i][j] = 'g';
                    nd.x = i,nd.y = j;
                    q.push(nd);
                }
            }
        }
    }
}

class GooseInZooDivTwo
{
        public:

        int count(vector <string> fd, int dis)
        {
            int i,j;
            n = fd.size(); m = fd[0].size();
            tmp.clear();
            for (i = 0; i < n; ++i)
            {
                tmp.push_back(fd[i]);
            }
            n = fd.size(); m = fd[0].size();
            ll cnt = 0;
            for (i = 0;  i < n; ++i)
            {
                for (j = 0; j < m; ++j)
                {
                    if (tmp[i][j] == 'v')
                    {
                        cnt++;
                        tmp[i][j] = 'g';
                        bfs(i,j,dis);
                    }
                }
            }
//            cout<<cnt<<endl;
            ll ans = 1;
            for (i = 0; i < cnt; ++i)
            {
                ans = ans*2%mod;
            }
            return ans - 1;
        }


};


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1000:

题意：

个顶n个段，每个段里面要么有狼，要么没狼，然后给出m个区域间，[l[i],r[i]]表示在区间内至少有一只狼，问满足条件的一共有多少种可能数？

思路：

记忆化搜索DP , dp[i][j]表示到i个段，有j个区间还没有处理， 对于第i个段我们有两种选择要么方狼，要么不放狼。 如果方狼的话，dp[i][j] = dp[i - 1][k] k表示必须多少个区间被处理了。

如果不放的话，要必须满足区间条件。 dp[i][j] += dp[i - 1][j] 

View Code

#line 5 "WolfInZooDivTwo.cpp"
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll long long
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);
#define tr(container, it) for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)

#define M 107
#define N 307
using namespace std;

const ll mod = 1000000007;

struct node
{
    int l,r;
}nd[N];

ll pow2[N];
ll dp[N][N];
int off[N],cnt;

ll DP(int x,int y)
{
     int i;
     if (x < 0) return 1;
     if (dp[x][y] != -1) return dp[x][y];
     if (y == 0) return (dp[x][y] = pow2[x + 1]);

     dp[x][y] = 0;
     for (i = y; i >= 1; --i)
     {
         if (x > nd[i].r) break;
     }
     dp[x][y] += DP(x - 1,i);
     dp[x][y] %= mod;

     for (i = y; i >= 1; --i)
     {
         if (x == nd[i].l) break;
     }
     if (i == 0) dp[x][y] += DP(x - 1, y);

     dp[x][y] %= mod;
     return dp[x][y];
}
vector<int> getR(string s)
{
    istringstream is(s);
    int no;
    vector<int> rs;
    while (is >> no)
    {
        rs.push_back(no);
    }
    return rs;
}
class WolfInZooDivTwo
{
        public:
        void init()
        {
            pow2[0] = 1;
            for (int i = 1; i <= 300; ++i)
            {
                pow2[i] = pow2[i - 1]*2%mod;
            }
        }
        int count(int n, vector <string> L, vector <string> R)
        {
            int i;
            init();
            string lstr = "",rstr = "";
            for (size_t i = 0; i < L.size(); ++i) lstr += L[i];
            for (size_t i = 0; i < R.size(); ++i) rstr += R[i];

            vector<int> Ls = getR(lstr);
            vector<int> Rs = getR(rstr);
            if (Ls.size() != Rs.size()) puts("BUBIUBYUGB");
            set<int> sr(Rs.begin(),Rs.end());

            CL(off,-1);
            int m = Ls.size();
            for (i = 0; i < m; ++i) off[Rs[i]] = max(off[Rs[i]],Ls[i]);
            cnt = 0;
            set<int>::iterator it = sr.begin();
            for (; it != sr.end(); ++it)
            {
                nd[++cnt].l = off[*it];
                nd[cnt].r = *it;
            }
            CL(dp,-1);
            return DP(n - 1,cnt);
        }
};