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线段树区间更新维护最小值。。。记得下放标记。。。
如果线段树上的一个完整区间被修改,那么最小值和最大值增加相应的值后不变,
会改变是因为一部分改变而另外一部分没有改变所以维护一下就好。
询问的时候也要记得下放标记。。。
数据结构快忘了,贴个板。
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 2e5+1; struct Seg { ll lazy; ll Min; }tr[maxn<<2]; int a[maxn]; int n; #define lid (id<<1) #define rid (id<<1|1) void build(int l = 0,int r = n-1,int id = 1) { if(l == r) { tr[id].Min = a[l]; return; }//tr[rt].lazy } int mid = (l+r)>>1, lc = lid, rc = rid; build(l,mid,lc); build(mid+1,r,rc); tr[id].Min = min(tr[lc].Min,tr[rc].Min); } int ql,qr; ll val; #define Modify(id,v) tr[id].lazy += v; tr[id].Min += v; void updata(int l = 0,int r = n-1,int id = 1) { if(ql<=l&&r<=qr) { Modify(id,val) return; } int mid = (l+r)>>1, lc = lid, rc = rid; if(tr[id].lazy){ ll &t = tr[id].lazy; Modify(lc,t) Modify(rc,t) t = 0; } if(ql<=mid){ updata(l,mid,lc); } if(qr>mid) { updata(mid+1,r,rc); } tr[id].Min = min(tr[lc].Min,tr[rc].Min); } const ll INF = 0x3f3f3f3f3f3f3f3f; ll query(int l = 0,int r = n-1,int id = 1) { if(ql<=l&&r<=qr){ return tr[id].Min; } int mid = (l+r)>>1,lc = lid, rc = rid; if(tr[id].lazy){ ll &t = tr[id].lazy; Modify(lc,t); Modify(rc,t); t = 0; } ll ret = INF; if(ql<=mid){ ret = min(ret,query(l,mid,lc)); } if(qr>mid){ ret = min(ret,query(mid+1,r,rc)); } return ret; } bool sscan_l(int &x,char *&s) { while(!isdigit(*s) && *s != ‘-‘){ if(!*s) return false; s++; } bool fg; if(*s == ‘-‘) fg = true,x = 0; else x = *s-‘0‘,fg = false; while(s++,isdigit(*s)) x = x*10+*s-‘0‘; if(fg) x = -x; return true; } char Line[666]; int main() { //freopen("in.txt","r",stdin); scanf("%d",&n); for(int i = 0; i < n; i++) scanf("%d",a+i); build(); int m; scanf("%d\n",&m); while(m--){ gets(Line); char *p = Line; int num[3], i; for(i = 0; i < 3; i++ ){ if(!sscan_l(num[i],p)) break; } if(i == 2){ int l = num[0],r = num[1]; if(l<=r) { ql = l; qr = r; printf("%I64d\n",query()); }else { ql = 0; qr = r; ll ans = query(); ql = l; qr = n-1; ans = min(ans,query()); printf("%I64d\n",ans); } }else { int l = num[0], r = num[1]; val = num[2]; if(l<=r){ ql = l; qr = r; updata(); }else { ql = 0; qr = r; updata(); ql = l; qr = n-1; updata(); } } } return 0; }
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原文地址:http://www.cnblogs.com/jerryRey/p/4783779.html