hdu 5122 树状数组？..其实是乱搞

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http://acm.hdu.edu.cn/showproblem.php?pid=5122

求一个序列最少进行几次变换(把某个数一冒泡排序的方式改变位置)才能使其有序。

对于每一个数只要后面的数有比它小就要换
所以逆着记录最小值，计算下就可以了。。

但是一开始没想到...写了个树状数组过的..蠢哭了

<span style="font-size:14px;">#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <queue>#include <map>#include <iostream>#include <sstream>#include <algorithm>using namespace std;#define RD(x) scanf("%d",&x)#define RD2(x,y) scanf("%d%d",&x,&y)#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define clr0(x) memset(x,0,sizeof(x))#define clr1(x) memset(x,-1,sizeof(x))#define eps 1e-9const double pi = acos(-1.0);typedef long long LL;const int inf = 1000000000;const int maxn = 1e6+5;map<int,int > hash;//#define a first//#define b second//typedef pair<int,int> p2;//p2 s[maxn];int bit[maxn],b[maxn];template<class T> T min(T a,T b,T c){    return min(min(a,b),c);}int n;void add(int x,int y){    for(int i = x;i <= n;i += i&-i)        bit[i] += y;}int cal(int x){    int sum = 0;    for(int i = x;i > 0;i -= i&-i)        sum += bit[i];    return sum;}int work(){    clr0(bit);    RD(n);    int res = 0,ans = 0;    for(int i = 1;i <= n;++i){        RD(b[i]);    }    for(int i = n;i >= 1;--i){        if(0 != cal(b[i]))            ans++;        add(b[i],1);    }    return ans;}int main(){    int _,cas = 1;    RD(_);    while(_--){        printf("Case #%d: %d\n",cas++,work());    }    return 0;}/**/</span>

正解..时间上没差多少..

<span style="font-size:14px;">#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <queue>#include <map>#include <iostream>#include <sstream>#include <algorithm>using namespace std;#define RD(x) scanf("%d",&x)#define RD2(x,y) scanf("%d%d",&x,&y)#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define clr0(x) memset(x,0,sizeof(x))#define clr1(x) memset(x,-1,sizeof(x))#define eps 1e-9const double pi = acos(-1.0);typedef long long LL;const int inf = 1000000000;const int maxn = 1e6+5;map<int,int > hash;//#define a first//#define b second//typedef pair<int,int> p2;//p2 s[maxn];int bit[maxn],b[maxn],n;int work(){    RD(n);    int mx = 0,ans = 0;    for(int i = 1;i <= n;++i){        RD(b[i]);        mx = max(mx,b[i]);    }    for(int i = n;i >= 1;--i){        if(mx < b[i])            ans++;        mx = min(b[i],mx);    }    return ans;}int main(){    int _,cas = 1;    RD(_);    while(_--){        printf("Case #%d: %d\n",cas++,work());    }    return 0;}/**/</span>

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