hdu5542 树状数组优化dp

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南阳比赛的题目，队友一发AC，我也来试试

题意很容易想到n^3的动态规划，dp[i][j]表示到i位置取j长度的种类，这样dp[i][j] = sum( dp[k][j-1], iff a[k] < a[i], 0<=k<i)。基于这样的思路，把数据离散化一下，对于每个j建立一个树状数组，这样就可以一次求的sum。时间复杂度变成n^2 logn，3400ms AC也是比较狠

#include <iostream>#include <map>#include <set>#include <vector>#include <cstring>#include <cstdio>#include <climits>#define N 1005#define MOD 1000000007using namespace std;int a[N], dp[N][N], num;int t[N][N];inline int lowbit(int x) {    return x & (-x);}void add(int i, int j, int c) {    while(j < num) {        t[i][j] += c;        if( t[i][j] >= MOD) t[i][j] %= MOD;        j += lowbit(j);    }    return;}int sum(int i, int j) {    int re = 0;    while(j) {        re += t[i][j];        if(re >= MOD) re %= MOD;        j -= lowbit(j);    }    return re;}int main(int argc, char* argv[]) {    int n, m;    int tt, ca = 1;    scanf("%d", &tt);    while( tt-- ) {        scanf("%d%d", &n, &m);        set<int> s;        map<int, int> mp;        for(int i=0; i<n; ++i) {            scanf("%d", &a[i]);            s.insert(a[i]);        }        num = 1;        for(set<int>::iterator it = s.begin(); it != s.end(); ++it) {            mp[ *it ] = num++;        }        memset(dp, 0, sizeof(dp));        memset(t, 0, sizeof(t));        for(int i=0; i<n; ++i)        for(int j=1; j<=m; ++j) {            if( j == 1)                dp[i][j] = 1;            else                dp[i][j] += sum(j-1, mp[ a[i] ]-1);            if(dp[i][j] >= MOD) dp[i][j] %= MOD;        //    printf("%d %d %d\n", i, j, dp[i][j]);            add(j, mp[ a[i] ], dp[i][j]);        }        int re = 0;        for(int i=0; i<n; ++i) {            re += dp[i][m];            if(re >= MOD) re %= MOD;        }        printf("Case #%d: %d\n", ca++, re);    }    return 0;}

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