CF52C：线段树（区间更新，区间查询最小值）

CF52C

题解：很简单的一道题目，注意区间[x,y]x大于y，要分两段。

代码：

#include <cstring>
#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;
ll const inf = 1e15;
int const N = 200000 + 10;
ll n,q;
struct Node
{
	ll l,r;
	ll lazy,mind;
	void updata(ll val){
		lazy += val;
		mind += val;
	}
}node[N<<2];
void push_up(ll id){
	node[id].mind = min(node[id<<1].mind,node[id<<1|1].mind);
}
void push_down(ll id){
	ll lazy =node[id].lazy;
	if(lazy){
		node[id<<1].updata(lazy);
		node[id<<1|1].updata(lazy);
		node[id].lazy = 0;
	} 
}
void build(ll id,ll l,ll r){
	node[id].l = l,node[id].r = r;
	node[id].lazy = 0;
	if(l == r){
		scanf("%I64d",&node[id].mind);
	}
	else{
		ll mid = (l + r) >> 1;
		build(id<<1,l,mid);
		build(id<<1|1,mid+1,r);
		push_up(id);
	}
}
ll query(ll id,ll L,ll R){
	ll l = node[id].l,r = node[id].r;
	if(L <= l && r <= R)
		return node[id].mind;
	else{
		push_down(id);
		ll mid = (l + r) >> 1;
		ll res = inf;
		if(L <= mid)	res = min(res,query(id<<1,L,R));
		if(mid < R)		res = min(res,query(id<<1|1,L,R));
		return res;
	}
} 
void updata(ll id,ll L,ll R,ll k){   //将[l,r]区间的每个数加上k
	ll l = node[id].l,	r = node[id].r;
	if(L <= l && r <= R){
		node[id].updata(k);
	}else{ 
		ll mid = (l + r) >> 1;
		push_down(id);
		if(L <= mid)	updata(id<<1,L,R,k);
		if(mid < R)		updata(id<<1|1,L,R,k);
		push_up(id);
	}
}
int main(){
	scanf("%I64d",&n);
	build(1,1,n);
	ll x,y,z;
	char t;
	scanf("%d",&q);
	for(ll i=1;i<=q;i++){
		scanf("%I64d%I64d",&x,&y);
		x++,y++;
		t = getchar();
		if(t == ' '){  
			scanf("%I64d",&z);
			if(x > y){
				updata(1,x,n,z);
				updata(1,1,y,z);
			}else updata(1,x,y,z);
		}else{
			ll mind;
			if(x <= y)	mind = query(1,x,y);
			else	mind = min(query(1,x,n),query(1,1,y));
			printf("%I64d\n",mind);
		}
	}
	return 0;
}