HDU 1698 Just a Hook （线段树延迟标记（lazy））

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题意：有n个数初始值都为1，m个操作a，b，c，表示把区间[a,b]变为c，求最后n个数的和。

经典区间更新求和问题，需要用到延迟标记（或者说是懒惰标记），简单老说就是每次更新

的时候不要更新到底，用延迟标记使得更新延迟到下次需要更新或询问的时候。

#include<cstdio>#include<stdlib.h>#include<string.h>#include<string>#include<map>#include<cmath>#include<iostream>#include <queue>#include <stack>#include<algorithm>#include<set>using namespace std;#define INF 1e8#define eps 1e-8#define LL long long #define maxn 200005#define mod  1000000009struct Tree{int l,r,sum,lazy;}tree[maxn<<2];void PushUp(int rt){tree[rt].sum=tree[rt<<1].sum+tree[rt<<1|1].sum;}void PushDown(int rt){if(tree[rt].lazy){tree[rt<<1].lazy=tree[rt<<1|1].lazy=tree[rt].lazy;tree[rt<<1].sum=(tree[rt<<1].r-tree[rt<<1].l+1)*tree[rt].lazy;tree[rt<<1|1].sum=(tree[rt<<1|1].r-tree[rt<<1|1].l+1)*tree[rt].lazy;tree[rt].lazy=0;}}void build(int rt,int l,int r){tree[rt].l=l;tree[rt].r=r;tree[rt].sum=0;tree[rt].lazy=0;if(l==r){tree[rt].sum=1;return ;}int mid=(l+r)>>1;build(rt*2,l,mid);build(rt*2+1,mid+1,r);PushUp(rt);}void update(int rt,int l,int r,int v){if(tree[rt].l>=l&&tree[rt].r<=r){tree[rt].lazy=v;tree[rt].sum=(tree[rt].r-tree[rt].l+1)*v;return;}PushDown(rt);int mid=(tree[rt].l+tree[rt].r)>>1;if(mid>=r)update(rt*2,l,r,v);else if(mid<l)update(rt*2+1,l,r,v);else{update(rt*2,l,mid,v);update(rt*2+1,mid+1,r,v);}PushUp(rt);}int main(){int t,ca=1;scanf("%d",&t);while(t--){int n,m;scanf("%d%d",&n,&m);build(1,1,n);int a,b,c;while(m--){scanf("%d%d%d",&a,&b,&c);update(1,a,b,c);}printf("Case %d: The total value of the hook is %d.\n",ca++,tree[1].sum);}return 0;}

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