UVaLive 6801 Sequence (计数DP)

题意：给定一个序列，有 n 个数，只有01，然后你进行k次操作，把所有的1变成0，求有多种方法。

析：DP是很明显的，dp[i][j] 表示进行第 i 次操作，剩下 j 个1，然后操作就两种，把1变成0，把0变成1。也可以用记忆化来做。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define frer freopen("in.txt", "r", stdin)
#define frew freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {1, 0, -1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL dp[maxn][maxn];

int main(){
    int T;  cin >> T;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%d %d", &n, &m);
        memset(dp, 0, sizeof dp);
        int x, cnt = 0;
        for(int i = 0; i < n; ++i){
            scanf("%d", &x);
            cnt += x;
        }
        dp[0][cnt] = 1;
        for(int i = 1; i <= m; ++i){
            for(int j = 0; j <= n; ++j){
                if(j > 0)    dp[i][j] = (dp[i][j] + dp[i-1][j-1] * (n-j+1)) % mod;
                if(j < n)  dp[i][j] = (dp[i][j] + dp[i-1][j+1] * (j+1)) % mod;
            }
        }
        printf("Case #%d: %lld
", kase, dp[m][0]);
    }
    return 0;
}

　　

记忆化搜索：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define frer freopen("in.txt", "r", stdin)
#define frew freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {1, 0, -1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL dp[maxn][maxn];

LL dfs(int k, int cnt){
    if(k == 0)  return cnt == 0;
    LL &ans = dp[k][cnt];
    if(ans >= 0)  return ans;
    int cnt0 = n - cnt;
    ans = 0;
    if(cnt > 0)  ans = (ans + dfs(k-1, cnt-1) * cnt) % mod;
    if(cnt0 > 0) ans = (ans + dfs(k-1, cnt+1) * cnt0) % mod;
    return ans;
}

int main(){
    int T;  cin >> T;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%d %d", &n, &m);
        memset(dp, -1, sizeof dp);
        int x, cnt = 0;
        for(int i = 0; i < n; ++i){
            scanf("%d", &x);
            cnt += x;
        }
        printf("Case #%d: %lld
", kase, dfs(m, cnt));
    }
    return 0;
}