NYOJ 119 士兵杀敌（三）【ST算法】

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题目链接：http://acm.nyist.net/JudgeOnline/problem.php?pid=119

解题思路：

RMQ算法。

不会的可以去看看我总结的RMQ算法。

http://blog.csdn.net/niushuai666/article/details/6624672

代码如下：

#include<cstdio>#include<algorithm>#include<cmath>using namespace std;const int N = 100010;int maxsum[N][20], minsum[N][20];void RMQ(int num) //预处理->O(nlogn){for(int j = 1; j < 20; ++j)for(int i = 1; i <= num; ++i)if(i + (1 << j) - 1 <= num){maxsum[i][j] = max(maxsum[i][j - 1], maxsum[i + (1 << (j - 1))][j - 1]);minsum[i][j] = min(minsum[i][j - 1], minsum[i + (1 << (j - 1))][j - 1]);}}int main(){int num, query;int src, des;scanf("%d %d", &num, &query);for(int i = 1; i <= num; ++i) //输入信息处理{scanf("%d", &maxsum[i][0]);minsum[i][0] = maxsum[i][0];}RMQ(num);while(query--) //O(1)查询{scanf("%d %d", &src, &des);int k = (int)(log(des - src + 1.0) / log(2.0));int maxres = max(maxsum[src][k], maxsum[des - (1 << k) + 1][k]);int minres = min(minsum[src][k], minsum[des - (1 << k) + 1][k]);printf("%d\n", maxres - minres);}return 0;}

代码优化后：

#include<cstdio>#include<algorithm>#include<cmath>using namespace std;const int N = 100010;int maxsum[20][N], minsum[20][N]; //优化1void RMQ(int num) //预处理->O(nlogn){for(int i = 1; i != 20; ++i)for(int j = 1; j <= num; ++j)if(j + (1 << i) - 1 <= num){maxsum[i][j] = max(maxsum[i - 1][j], maxsum[i - 1][j + (1 << i >> 1)]); //优化2minsum[i][j] = min(minsum[i - 1][j], minsum[i - 1][j + (1 << i >> 1)]);}}int main(){int num, query;int src, des;scanf("%d %d", &num, &query);for(int i = 1; i <= num; ++i) //输入信息处理{scanf("%d", &maxsum[0][i]);minsum[0][i] = maxsum[0][i];}RMQ(num);while(query--) //O(1)查询{scanf("%d %d", &src, &des);int k = (int)(log(des - src + 1.0) / log(2.0));int maxres = max(maxsum[k][src], maxsum[k][des - (1 << k) + 1]);int minres = min(minsum[k][src], minsum[k][des - (1 << k) + 1]);printf("%d\n", maxres - minres);}return 0;}

优化1：数组由F[N][20]变为F[20][N];

因为数组的地址为a + i + j，对应上面数组，我们需要先循环N的部分，所以

如果是第一种，我们计算时因为i不断变化，我们就需要计算a + i + j

如果是第二种，我们计算时a + i不变,只需要改变j

优化2：

位运算

转载至：http://blog.csdn.net/niushuai666/article/details/7400587

自己也放心不下ST算法

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