hdu 5273 Dylans loves sequence 逆序数 区间dp

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题意：给n个数，q次询问，（L，R）区间内的逆序数。

思路： 区间dp

代码一：

#include <bits/stdc++.h>using namespace std;typedef long long ll;const int maxn = 1e3+10;const int INF = 0x3f3f3f3f;const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;inline ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}//////////////////////////////////////////////////////////////////////////int a[maxn],dp[maxn][maxn];int main(){int n,q; n = read(), q = read();for(int i=1; i<=n; i++)a[i] = read();for(int len=1; len<n; len++)for(int i=1; i+len<=n; i++){int j = i+len;dp[i][j] = dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1];if(a[i] > a[j])dp[i][j]++;}for(int i=0; i<q; i++){int l,r; l=read(),r=read();cout << dp[l][r] << endl;}    return 0;}

思路二：

dp[i][j], 先求出每个i为起始位置的逆序数， dp[i][j] = dp[i][j-1];

再移动i，求出任意（L，R）区间内的逆序数。 dp[i][j] = dp[i+1][j];

代码二：

#include <bits/stdc++.h>using namespace std;typedef long long ll;const int maxn = 1e3+10;const int INF = 0x3f3f3f3f;const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;inline ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}//////////////////////////////////////////////////////////////////////////int a[maxn],dp[maxn][maxn];int main(){int n,q; n=read(),q=read();for(int i=1; i<=n; i++)a[i] = read();for(int i=1; i<=n; i++){ //dp[i][j]是[i,j]区间里i为起始位置的倒置数对for(int j=i+1; j<=n; j++)if(a[i] > a[j]){dp[i][j]++;// cout << "111dp[" << i << "][" << j << "] = " << dp[i][j] << endl;}for(int j=i+1; j<=n; j++){dp[i][j] += dp[i][j-1];// cout << "222dp[" << i << "][" << j << "] = " << dp[i][j] << endl;}}for(int i=n-1; i>=1; i--) //再枚举[i,j]这个区间里面任意一个数为起始位置，含有的倒置数对for(int j=i+1; j<=n; j++){dp[i][j] += dp[i+1][j];// cout << "333dp[" << i << "][" << j << "] = " << dp[i][j] << endl;}while(q--){int l,r; l=read(),r=read();cout << dp[l][r] << endl;}    return 0;}