• HDU


    https://cn.vjudge.net/problem/HDU-1698

    题意

    大小为n的数组,数组元素初始值为1,有q次操作,x,y,z表示从第x到第y所有的元素的值变为z,最后问1到n的和。

    分析

    区间修改,给每个区间打标记。注意这里是直接把整个区间都变为某个数。

    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <string>
    #include <algorithm>
    #include <cmath>
    #include <ctime>
    #include <vector>
    #include <queue>
    #include <map>
    #include <stack>
    #include <set>
    #include <bitset>
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    #define ms(a, b) memset(a, b, sizeof(a))
    #define pb push_back
    #define mp make_pair
    #define pii pair<int, int>
    #define eps 0.0000000001
    #define IOS ios::sync_with_stdio(0);cin.tie(0);
    #define random(a, b) rand()*rand()%(b-a+1)+a
    #define pi acos(-1)
    const ll INF = 0x3f3f3f3f3f3f3f3fll;
    const int inf = 0x3f3f3f3f;
    const int maxn = 100000 + 10;
    const int maxm = 200000 + 10;
    const int mod = 998244353;
    int n;
    struct ND{
        int l,r;
        ll sum,lazy;
    }tree[maxn<<2];
    int a[maxn];
    void pushup(int rt){
        tree[rt].sum=tree[rt<<1].sum+tree[rt<<1|1].sum;
    }
    void pushdown(int rt,int len){
        if(tree[rt].lazy){
            tree[rt<<1].lazy=tree[rt].lazy;
            tree[rt<<1|1].lazy=tree[rt].lazy;
            tree[rt<<1].sum=tree[rt].lazy*(len-(len>>1));
            tree[rt<<1|1].sum=tree[rt].lazy*(len>>1);
            tree[rt].lazy=0;
        }
    }
    void build(int rt,int l,int r){
        tree[rt].l=l,tree[rt].r=r;
        tree[rt].lazy=0;
        if(l==r){
            tree[rt].sum=1;
            return;
        }
        int mid=(l+r)>>1;
        build(rt<<1,l,mid);
        build(rt<<1|1,mid+1,r);
        pushup(rt);
    }
    void update(int rt,int L,int R,int val){
        if(L<=tree[rt].l&&tree[rt].r<=R){
            tree[rt].lazy=val;
            tree[rt].sum=1ll*val*(tree[rt].r-tree[rt].l+1);
            return;
        }
        pushdown(rt,tree[rt].r-tree[rt].l+1);
        int mid=(tree[rt].l+tree[rt].r)>>1;
        if(mid>=L) update(rt<<1,L,R,val);
        if(mid<R) update(rt<<1|1,L,R,val);
        pushup(rt);
    }
    ll query(int rt,int L,int R){
        if(L<=tree[rt].l&&tree[rt].r<=R) return tree[rt].sum;
        pushdown(rt,tree[rt].r-tree[rt].l+1);
        ll sum=0;
        int mid=(tree[rt].l+tree[rt].r)>>1;
        if(mid<R) sum+=query(rt<<1|1,L,R);
        if(mid>=L) sum+=query(rt<<1,L,R);
        return sum;
    }
    int main() {
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
    //    freopen("output.txt", "w", stdout);
    #endif
        int t;
        int cas=1;
        scanf("%d",&t);
        while(t--){
            scanf("%d",&n);
    //        for(int i=1;i<=n;i++) a[i]=1;
            build(1,1,n);
            int q;
            scanf("%d",&q);
            while(q--){
                int x,y,z;
                scanf("%d%d%d",&x,&y,&z);
                update(1,x,y,z);
            }
            printf("Case %d: The total value of the hook is %lld.
    ",cas++,query(1,1,n));
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/fht-litost/p/9572183.html
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