UVA12299 RMQ with Shifts 线段树查询 单点更新

线段树的题目，不是特别难，掌握单点更新 和 区间查找 即可，给一个数组，下标1到n,有一个shift操作，就是把它给你的下标 最前面一个放到最后面一个，相应位置的值也发生改变，然后有询问query(l,r)求出闭区间[l,r]的最小值，因为shift这一命令语句 题目说不超过30个字符，除去括号shift字符串 还有逗号，数字部分 最多也就13个，所以可以把这个强行用单点更新来做，有点暴力，

总是做算法，不如来个陶冶情操的文章一篇： http://www.sanwen.net/subject/3628849/

#include<iostream>
#include<cstdio>
#include<list>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<cmath>
#include<memory.h>
#include<set>

#define ll long long

#define eps 1e-8

#define inf 0xfffffff

//const ll INF = 1ll<<61;

using namespace std;

//vector<pair<int,int> > G;
//typedef pair<int,int > P;
//vector<pair<int,int> > ::iterator iter;
//
//map<ll,int >mp;
//map<ll,int >::iterator p;

const int N = 100000 + 5;

typedef struct Node {
	int l,r,minn;
};

Node tree[N * 6];

int a[N];
int Stack[N];

void clear() {
	memset(tree,0,sizeof(tree));
	memset(a,0,sizeof(a));
	memset(Stack,0,sizeof(Stack));
}

void build(int l,int r,int id) {
	tree[id].l = l;
	tree[id].r = r;
	if(l == r) {
		tree[id].minn = a[l];
		return;
	}
	int mid = (l + r) / 2;
	build(l,mid,id*2);
	build(mid+1,r,id*2 + 1);
	tree[id].minn = min(tree[id*2].minn,tree[id*2 + 1].minn);
}

int find(int l,int r,int id) {
	if(l == tree[id].l && r == tree[id].r)
		return tree[id].minn;
	int mid = (tree[id].l + tree[id].r)/2;
	if(r <= mid)return find(l,r,id*2);
	else if(mid < l)return find(l,r,id*2 + 1);
	return min(find(l,mid,id<<1),find(mid+1,r,id*2 + 1));
}

void update(int pos,int val,int id) {
	if(tree[id].l == tree[id].r) {
		tree[id].minn = val;
		return;
	}
	int mid = (tree[id].l + tree[id].r) / 2;
	if(pos <= mid)update(pos,val,id*2);
	else update(pos,val,id*2 + 1);
	tree[id].minn = min(tree[id*2].minn,tree[id*2 + 1].minn);
}

int main() {
	int n,q;
	while(scanf("%d %d",&n,&q) ==2 ) {
		clear();
		for(int i=1;i<=n;i++)
			scanf("%d",&a[i]);
		build(1,n,1);
		while(q--) {
			char c = ‘*‘;
			int u,v;
			while(c != ‘q‘ && c!= ‘s‘) c = getchar();
			int mark = 5;
			while(mark--)getchar();
			if(c == ‘q‘) {
				scanf("%d,%d",&u,&v);
				printf("%d\n",find(u,v,1));
				continue;
			}
			int top = 0;
			while(true) {
				scanf("%d",&u);
				Stack[top++] = u;
				c = getchar();
				if(c == ‘)‘)break;
			}
			for(int i=0;i<top;i++)
				update(Stack[i],a[Stack[(i+1)%top]],1);
			v = a[Stack[0]];
			for(int i=0;i<top-1;i++)
				a[Stack[i]] = a[Stack[i+1]];
			a[Stack[top-1]] = v;
		}
	}
	return EXIT_SUCCESS;
}

UVA12299 RMQ with Shifts 线段树查询 单点更新,布布扣,bubuko.com

UVA12299 RMQ with Shifts 线段树查询 单点更新

原文：http://blog.csdn.net/yitiaodacaidog/article/details/21882785