离线+dp 51Nod1020 逆序排列

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题意：求长度为n的排列的逆序对等于k的排列总数，取模1e9+7

思路：刚开始想找规律，但是发现根本没法搞。。

但是我们如果从dp的角度去思考，就能找到一个dp是这样的

dp[n][k]=dp[n-1][i],max(k-n+1,0)<=i<=k

其中边界dp[0][0]=1

那么就能用离线+前缀和去搞答案了

要注意的就是ans等于0的情况，离线的时候要注意p是否能取到所有的，不然可能后面的答案会全错

#include<map>#include<set>#include<cmath>#include<ctime>#include<stack>#include<queue>#include<cstdio>#include<cctype>#include<string>#include<vector>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define fuck(x) cout<<"["<<x<<"]"#define FIN freopen("input.txt","r",stdin)#define FOUT freopen("output.txt","w+",stdout)using namespace std;typedef long long LL;typedef pair<int, int>PII;const int MX = 2e4 + 5;const int mod = 1e9 + 7;const int INF = 0x3f3f3f3f;struct Data {    int id, n, k;    bool operator<(const Data &P) const {        if(n == P.n) return k < P.k;        return n < P.n;    }} A[MX];int n, ans[MX];LL dp[2][MX];int main() {    //FIN;    while(~scanf("%d", &n)) {        memset(ans, 0, sizeof(ans));        int mn = 0, mk = 0;        for(int i = 1; i <= n; i++) {            A[i].id = i;            scanf("%d%d", &A[i].n, &A[i].k);            mn = max(mn, A[i].n);            mk = max(mk, A[i].k);        }        sort(A + 1, A + 1 + n);        int cur = 0, nxt = 1, p = 1;        for(int i = 0; i <= mk; i++) dp[cur][i] = 1;        for(int i = 1; i <= mn; i++) {            int bound = i * (i - 1) / 2;            for(int j = 0; j <= min(mk, bound); j++) {                int l = max(j - i + 1, 0), r = j;                LL sl = l == 0 ? 0 : dp[cur][l - 1];                dp[nxt][j] = (dp[cur][r] - sl + mod) % mod;                while(p <= n && i == A[p].n && j == A[p].k) {                    ans[A[p++].id] = dp[nxt][j];                }                dp[nxt][j] = (dp[nxt][j] + dp[nxt][j - 1]) % mod;            }            for(int j = min(mk, bound) + + 1; j <= mk; j++) {                dp[nxt][j] = dp[nxt][j - 1];                while(p <= n && i == A[p].n && j == A[p].k) {                    ans[A[p++].id] = 0;                }            }            swap(cur, nxt);        }        for(int i = 1; i <= n; i++) {            printf("%d\n", ans[i]);        }    }    return 0;}

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