题意:有t组测试数据,n为钩子长度(1<=n<=100000),m为操作的次数。初始时,每个钩子的价值为1,操作由三个数字组成x,y,z表示把区间[x,y]的钩子变成的价值变成z(1代表铜,2银,3金)。
使用了延迟标记type,表示当前区间的左右子儿子的价值为type,在更新到当前区间时,同时更新当前区间的valu,即type*当前区间的长度。
/*代码风格更新后*/ #include <iostream> #include <cstdio> #include <cstring> using namespace std; #define LL(x) (x<<1) #define RR(x) (x<<1|1) #define MID(a,b) (a+((b-a)>>1)) const int N=100005; struct node { int lft,rht; int valu,type; int mid(){return MID(lft,rht);} void fun(int tmp) { type=tmp; valu=(rht-lft+1)*type; } }; struct Segtree { node tree[N*4]; void build(int lft,int rht,int ind) { tree[ind].lft=lft; tree[ind].rht=rht; tree[ind].valu=rht-lft+1; tree[ind].type=1; if(lft!=rht) { int mid=tree[ind].mid(); build(lft,mid,LL(ind)); build(mid+1,rht,RR(ind)); } } void updata(int st,int ed,int ind,int type) { int lft=tree[ind].lft,rht=tree[ind].rht; if(st<=lft&&rht<=ed) tree[ind].fun(type); else { if(tree[ind].type) { tree[LL(ind)].fun(tree[ind].type); tree[RR(ind)].fun(tree[ind].type); tree[ind].type=0; } int mid=tree[ind].mid(); if(st<=mid) updata(st,ed,LL(ind),type); if(ed>mid) updata(st,ed,RR(ind),type); tree[ind].valu=tree[LL(ind)].valu+tree[RR(ind)].valu; } } }seg; int main() { int t,t_cnt=0; scanf("%d",&t); while(t--) { int n,m,a,b,c; scanf("%d%d",&n,&m); seg.build(1,n,1); while(m--) { scanf("%d%d%d",&a,&b,&c); seg.updata(a,b,1,c); } printf("Case %d: The total value of the hook is %d.\n",++t_cnt,seg.tree[1].valu); } return 0; }
/*代码风格更新前*/ #include <iostream> #include <cstdio> using namespace std; const int N=100005; struct node { int left,right,co,sum; int mid(){return left+(right-left)/2;} int dis(){return right-left+1;} void change(int a) { co=a; sum=co*dis(); } }; struct Segtree { node tree[N*4]; void build(int left,int right,int r) { tree[r].left=left; tree[r].right=right; tree[r].co=1; tree[r].sum=tree[r].dis(); if(left<right) { int mid=tree[r].mid(); build(left,mid,r*2); build(mid+1,right,r*2+1); } } void updata(int be,int end,int r,int co) { if(be<=tree[r].left&&tree[r].right<=end) { tree[r].change(co); } else { if(tree[r].co!=0) { tree[r*2].change(tree[r].co); tree[r*2+1].change(tree[r].co); tree[r].co=0; } int mid=tree[r].mid(); if(be<=mid) updata(be,end,r*2,co); if(end>mid) updata(be,end,r*2+1,co); tree[r].sum=tree[r*2].sum+tree[r*2+1].sum; } } }seg; int main() { int t,t_cnt=0; scanf("%d",&t); while(t--) { int n,m; scanf("%d%d",&n,&m); seg.build(1,n,1); for(int i=0;i<m;i++) { int x,y,z; scanf("%d%d%d",&x,&y,&z); seg.updata(x,y,1,z); } printf("Case %d: The total value of the hook is %d.\n",++t_cnt,seg.tree[1].sum); } return 0; }