bzoj2431

dp

设dp[i][j]为当前到了第i个位置,一共出现了j个逆序对,然后考虑转移,因为前面只有i-1个数,所以这位只能产生[0,i-1]对逆序对,那么dp[i][j]=sigma(dp[i-1][k]),k∈[i-j+1,j]。

但是这样直接搞是O(n^3)的,那么我们用前缀和优化一下,就是O(n^2)的了。

碰见这种排列的题不能直接把现在填的数设为状态,我们可以通过调整之前填的数使当前要放进去的数是合法的,所以之前的数就是不确定的,应该考虑用其他条件去设状态

#include<bits/stdc++.h>
using namespace std;
const int N = 1010; 
namespace IO 
{
    const int Maxlen = N * 50;
    char buf[Maxlen], *C = buf;
    int Len;
    inline void read_in()
    {
        Len = fread(C, 1, Maxlen, stdin);
        buf[Len] = '\0';
    }
    inline void fread(int &x) 
    {
        x = 0;
        int f = 1;
        while (*C < '0' || '9' < *C) { if(*C == '-') f = -1; ++C; }
        while ('0' <= *C && *C <= '9') x = (x << 1) + (x << 3) + *C - '0', ++C;
        x *= f;
    }
    inline void read(int &x)
    {
        x = 0;
        int f = 1; char c = getchar();
        while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
        while(c >= '0' && c <= '9') { x = (x << 1) + (x << 3) + c - '0'; c = getchar(); }
        x *= f;
    }
    inline void read(long long &x)
    {
        x = 0;
        long long f = 1; char c = getchar();
        while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
        while(c >= '0' && c <= '9') { x = (x << 1ll) + (x << 3ll) + c - '0'; c = getchar(); }
        x *= f;
    } 
} using namespace IO;
int n, k;
int dp[N][N];
int main()
{
    read(n);
    read(k);
    for(int i = 0; i <= k; ++i) dp[0][i] = 1;
    for(int i = 1; i <= n; ++i)
        for(int j = 0; j <= k; ++j)
        {
            dp[i][j] = dp[i - 1][j];
            if(j >= i) dp[i][j] = dp[i][j] - dp[i - 1][j - i];
            if(j) dp[i][j] = dp[i][j] + dp[i][j - 1];
            if(dp[i][j] >= 10000) dp[i][j] -= 10000;
            if(dp[i][j] < 0) dp[i][j] += 10000;
        }
    if(k == 0) puts("1");
    else printf("%d\n", ((dp[n][k] - dp[n][k - 1]) % 10000 + 10000) % 10000);   
    return 0;
} 
View Code

 

posted @ 2017-09-19 16:20  19992147  阅读(124)  评论(0编辑  收藏  举报