poj 3468<线段树区间求和，区间修改，树状数组模版>

poj 3468 题意就是区间修改和区间求和。

开始没考虑如对1-10这个区间只更新了1-5，求和1-10的时候没加这个更新，应该结构上加上个b。

#include<iostream>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <map>#include <set>#include <algorithm>using namespace std;const int maxn=100000+7;struct ttt{    long long l,r,w,a,b;};long long qq[maxn];ttt tree[4*maxn];int build(long long l,long long r,int num){    tree[num].l=l;    tree[num].r=r;    if(l==r){        tree[num].w=qq[l];return 0;    }    build(l,(l+r)/2,num+num);    build((l+r)/2+1,r,num+num+1);    tree[num].w=tree[num+num].w+tree[num+num+1].w;}int update(long long l,long long r,long long y,int num){    if(tree[num].l>=l&&tree[num].r<=r){ //区间属于l,r        tree[num].a+=y;        return 0;    }    tree[num].b+=y*(min(r,tree[num].r)-max(l,tree[num].l)+1);//考虑这个区间只加部分，但是选这个区间    //得考虑这部分和也得选上    int mid1=(tree[num].l+tree[num].r)/2;    if(l>mid1){        update(l,r,y,num+num+1);    }else if(r<=mid1){        update(l,r,y,num+num);    }else if(l<=mid1&&r>mid1){        update(l,r,y,num+num);        update(l,r,y,num+num+1);    }}long long sum1;int find1(long long l,long long r,int num){    if(l<=tree[num].l&&r>=tree[num].r){ //区间属于l,r        sum1+=tree[num].w+tree[num].a*(tree[num].r-tree[num].l+1);        sum1+=tree[num].b;    }else{        sum1+=tree[num].a*(min(r,tree[num].r)-max(l,tree[num].l)+1); //l,r在这个区间的值     int mid1=(tree[num].l+tree[num].r)/2;    if(l>mid1){        find1(l,r,num+num+1);    }else if(r<=mid1){        find1(l,r,num+num);    }else if(l<=mid1&&r>mid1){        find1(l,r,num+num);        find1(l,r,num+num+1);    }    }}int main(){long long i,j,k,f1,f2,f3,f4,t1,t2,t3,t4,n,m;    //reopen("in.txt","r",stdin);memset(tree,0,sizeof(tree));cin >> n >> m;t3=0;    for(i=1;i<=n;i++){        scanf("%lld",&qq[i]);    }    scanf("\n");    char b;    build(1,n,1);    for(i=1;i<=m;i++){        scanf("%c %lld %lld",&b,&t1,&t2);         if(b=='Q'){        scanf("\n");        sum1=0;        find1(t1,t2,1);        printf("%lld\n",sum1);        }else{         scanf("%lld\n",&t3);         update(t1,t2,t3,1);        }    }return 0;}

树状数组也能区间更新和查找，速度会更快一些，开的数组就是n的大小。

#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <vector>#include <algorithm>#include <set>using namespace std;const int maxn=1e6+7;typedef long long ll;ll n,c1[maxn],c2[maxn];ll lowbits(ll x){ return x&(-x);}void add(ll *r,ll pos,ll v){while(pos<=n){r[pos]+=v;pos+=lowbits(pos);}}ll sigma(ll *r,ll pos){ll anx=0;while(pos>0){anx+=r[pos];pos-=lowbits(pos);}return anx;}int main(){//freopen("in.txt","r",stdin);long long i,j,k,f1,f2,f3,t1,t2,t3,sum1,sum2;int m;cin >> n>>m;for(i=1;i<=n;i++){        scanf("%lld",&f3);        f1=f2=i;        add(c1,f1,f3);add(c1,f2+1,-f3);            add(c2,f1,f3*(f1-1));add(c2,f2+1,-f3*f2);}char b;scanf("\n");while(m--){scanf("%c %lld %lld",&b,&f1,&f2);if(b=='C'){ //t2==1时候,f1-f2加f3scanf("%lld\n",&f3);add(c1,f1,f3);add(c1,f2+1,-f3);            add(c2,f1,f3*(f1-1));add(c2,f2+1,-f3*f2);}else{ //t2==2 的时候 查询f1-f2的和scanf("\n");sum1=(f1-1)*sigma(c1,f1-1)-sigma(c2,f1-1);            sum2=f2*sigma(c1,f2)-sigma(c2,f2);            cout << sum2-sum1 << endl;}}return 0;}