Devour Magic
n个单位,每个单位每秒增加1法力,在某些时间取走一些区间的法力值(取走之后该区间所有单位的法力变为0),求取得的所有法力值。
就是在原来的基础上增加了清零的操作,不过这个清零(实际代码中也可以置为任意值)的操作通过flag标志和一个sset变量来保存下要置的数,其他操作和正常pushdown一样,每次在输入时记录上一次更新的时间last,这一次直接t-last就好了。。。
之前的一个超时代码也先粘在这里吧,(标记的可能多搜了!!!),但是提交AC的代码和自己的思路是一模一样的,具体现在也不知道问题出在哪里
超时代码:
///线段树区间更新 #include<algorithm> #include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #include<vector> #include<queue> #include<stack> #include<iomanip> #include<string> #include<climits> #include<cmath> #define INF 0x3f3f3f3f #define MAX 1100000 #define LL long long using namespace std; LL n,m; LL coun=0; int xx; struct Tree { LL l,r; LL sum,add; }; Tree tree[MAX*8]; void pushup(LL x) ///更新父节点 { tree[x].sum=tree[ x<<1 ].sum+tree[ (x<<1)+1 ].sum; } void pushdown(LL x) ///用于更新add数组 { if(x>xx){ tree[x].sum = 0; return ; } LL tmp = x<<1 ; if(tree[x].add == -INF){ // cout<<x<<"----------------------"<<endl; tree[tmp].add = -INF ; ///由子节点通过增加 tree[tmp+1].add = -INF; tree[tmp].sum = 0; tree[tmp+1].sum = 0; tree[x].add=0; tree[x].sum=0; return ; } if( tmp > xx ) return ; if(tree[tmp].add == -INF ) pushdown(tmp); if(tree[tmp+1].add == -INF) pushdown(tmp+1); tree[tmp].add += tree[x].add; ///由子节点通过增加 tree[tmp].sum += tree[x].add*(tree[tmp].r-tree[tmp].l+1); tree[tmp+1].add += tree[x].add; tree[tmp+1].sum += tree[x].add*(tree[tmp+1].r-tree[tmp+1].l+1); if(tmp>xx) cout<<tmp<<" "<<tree[tmp].add<<" "<<tree[tmp].add<<" "<<tree[tmp].l<<" "<<tree[tmp].r<<endl; tree[x].add=0; } void build(int l,int r,int x) { tree[x].l=l , tree[x].r=r , tree[x].add=0; if(l==r) { tree[x].sum = 0; ///子节点初始化 tree[x].add = 0; return ; } int tmp=x<<1; int mid=(l+r)>>1; build(l,mid,tmp); build(mid+1,r,tmp+1); pushup(x); ///建立时需根据子节点更细父亲节点 } void update(LL l,LL r,LL c,LL x) ///分别表示区间的左 , 右 , 增加的值 ,当前父亲节点 { if(r<tree[x].l||l>tree[x].r||x>xx) return ; if(l<=tree[x].l&&r>=tree[x].r) ///该区间为需要更新区间的子区间 { // cout<<"************\n"; if(c==-INF){ tree[x].add = -INF; tree[x].sum = 0; return ; } if(x>xx) return ; if(tree[x].add == -INF ) pushdown(x); tree[x].add += c ; tree[x].sum += c*(tree[x].r-tree[x].l+1); ///区间长度*要增加的数值 return ; } ///如果比当前的范围还小,就通过该节点向下更新下面的节点 if(tree[x].add ) pushdown(x); ///更新从上向下更新add LL tmp=x<<1; if(tmp>xx){ pushup(x); return ; } update(l,r,c,tmp); update(l,r,c,tmp+1); } LL query(LL l,LL r,LL x) { ///if(r<tree[x].l||l>tree[x].r) return -INF;//要更新的区间不在该区间上(输入有误) if(l<=tree[x].l&&r>=tree[x].r) ///要计算的区间包括了该区间 { return tree[x].sum; } if( tree[x].add&&tree[x].add!=-INF ) pushdown(x); ///如果add不为零,对查询可能用到的add更新 LL tmp=x<<1; if(tmp>xx) return 0; LL mid=(tree[x].l+tree[x].r)>>1; if(r<=mid) return query(l,r,tmp); else if(l>mid) return query(l,r,tmp+1); else return query(l,mid,tmp)+query(mid+1,r,tmp+1); } int main() { int xxx; scanf("%d",&xxx); while(xxx--) { scanf("%I64d%I64d",&n,&m); xx=4*n; coun=0; build(1,n,1); LL last=0; while(m--) { LL l,r,time; scanf("%I64d%I64d%I64d",&time,&l,&r); update(1,n,time-last,1); /* cout<<"加"<<time-last<<endl; for(int i=1;i<=60;i++) cout<<i<<":"<<tree[i].add<<' '<<tree[i].sum<<" "; cout<<endl;*/ last = time ; coun+=query(l,r,1); update(l,r,-INF,1); /* for(int i=1;i<=60;i++) cout<<i<<":"<<tree[i].add<<' '<<tree[i].sum<<" "; cout<<endl;*/ } printf("%I64d\n",coun); } return 0; } /************************************** Problem id : SDUT OJ 2880 User name : 张士卫 Result : Time Limit Exceeded Take Memory : 0K Take Time : 2010MS Submit Time : 2016-04-02 09:36:52 **************************************/
AC代码:
#include<iostream> #include<stdio.h> #include<string.h> using namespace std; #define L(root) ((root)<<1) #define R(root) (((root)<<1)|1) const int MAXN=1e5+10;// int numbers[MAXN];//初始值 struct node{ int left,right;/// long long sum; int add;///区间增加值 int sset; ///区间里的数设为v bool flag;///标记是否设置为v int mid(){ return ((right+left)>>1); } }tree[MAXN*4];//4倍空间 void pushUp(int root){ tree[root].sum=tree[L(root)].sum+tree[R(root)].sum; } void pushDown(int root){ int L=L(root),R=R(root); if(tree[root].flag){ tree[L].add=tree[R].add=0; tree[L].sset=tree[R].sset=tree[root].sset; tree[L].flag=tree[R].flag=true; tree[L].sum=tree[root].sset*(tree[L].right-tree[L].left+1); tree[R].sum=tree[root].sset*(tree[R].right-tree[R].left+1); tree[root].flag=false; } if(tree[root].add){ ///正常pushdown tree[L].add+=tree[root].add; tree[R].add+=tree[root].add; tree[L].sum+=tree[root].add*(tree[L].right-tree[L].left+1); tree[R].sum+=tree[root].add*(tree[R].right-tree[R].left+1); tree[root].add=0; } } void build(int root,int left,int right){ tree[root].left=left; tree[root].right=right; tree[root].add=0; tree[root].flag=false; if(left==right){ tree[root].sum=numbers[left]; return; } int mid=tree[root].mid(); build(L(root),left,mid); build(R(root),mid+1,right); pushUp(root); } long long query(int root,int left,int right){ if(tree[root].left==left&&tree[root].right==right) return tree[root].sum; pushDown(root); int mid=tree[root].mid(); if(right<=mid) return query(L(root),left,right); else if(left>mid) return query(R(root),left,right); else return query(L(root),left,mid)+query(R(root),mid+1,right); } void update(int root,int left,int right,int add){ if(tree[root].left==left&&tree[root].right==right){ tree[root].add+=add; tree[root].sum+=add*(right-left+1); return; } pushDown(root); int mid=tree[root].mid(),L=L(root),R = R(root); if(right<=mid) update(L,left,right,add); else if(left>mid) update(R,left,right,add); else{ update(L,left,mid,add); update(R,mid+1,right,add); } pushUp(root); } void setf(int root,int left,int right,int sset){ if(tree[root].left==left&&tree[root].right==right){ tree[root].add=0; tree[root].sum=sset*(right-left+1); tree[root].sset=sset; tree[root].flag=true; return; } pushDown(root); int mid=tree[root].mid(),L=L(root),R = R(root); if(right<=mid) setf(L,left,right,sset); else if(left>mid) setf(R,left,right,sset); else{ setf(L,left,mid,sset); setf(R,mid+1,right,sset); } pushUp(root); } int main(){ memset(numbers,0,sizeof(numbers)); int T; int n,m; int t,l,r; int i; int lastTime; long long sum; scanf("%d",&T); while(T--){ scanf("%d%d",&n,&m); build(1,1,n); lastTime=0; sum=0; for(i=0;i<m;++i){ scanf("%d%d%d",&t,&l,&r); update(1,1,n,t-lastTime); sum+=query(1,l,r); setf(1,l,r,0);//l到r区间设为0 lastTime=t; } printf("%lld\n",sum); } return 0; }