HDU 1698 线段树成段更新

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题意：给你T组数据，每组N个数，初始为1，M个操作，每个操作将区间a到b的值更新为才，问n个数的和。

思路：简单的线段树成段更新，需要用到懒惰标记，说白了懒惰标记就是用到的时候再去更新它，不然就放在那里

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>using namespace std;const int inf=0x3f3f3f3f;const int maxn=1e5+10;typedef long long ll;#define mm(a) memset(a,0,sizeof(a))int num[maxn<<2],num1[maxn<<2];void pushup(int node){    num[node]=num[node<<1]+num[node<<1|1];}void pushdown(int node,int node1){    if(num1[node]){        int t=num1[node];        num1[node<<1]=t;        num1[node<<1|1]=t;        num[node<<1]=(node1-(node1>>1))*t;        num[node<<1|1]=(node1>>1)*t;        num1[node]=0;    }}void buildtree(int le,int ri,int node){    if(le==ri){        num[node]=1;        return ;    }    int t=(le+ri)>>1;    buildtree(le,t,node<<1);    buildtree(t+1,ri,node<<1|1);    pushup(node);}void update(int l,int r,int add,int le,int ri,int node){    if(l<=le&&ri<=r){        num1[node]=add;        num[node]=add*(ri-le+1);        return ;    }    pushdown(node,ri-le+1);    int t=(le+ri)>>1;    if(l<=t) update(l,r,add,le,t,node<<1);    if(r>t) update(l,r,add,t+1,ri,node<<1|1);    pushup(node);}int main(){    int T,t=1,n,m;    scanf("%d",&T);    while(T--){        mm(num1);        scanf("%d%d",&n,&m);        int a,b,c;        buildtree(1,n,1);        while(m--){            scanf("%d%d%d",&a,&b,&c);            update(a,b,c,1,n,1);//            for(int i=1;i<=4*n;i++)//                cout<<num[i]<<" ";//            cout<<endl;        }        printf("Case %d: The total value of the hook is %d.\n",t++,num[1]);    }    return 0;}

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