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  • 论文补全计划

    https://github.com/OI-wiki/libs/blob/master/集训队历年论文/

    国家集训队2005论文集

    黄源河--左偏树的特点及其应用

    LuoguP4331 [BOI2004]Sequence 数字序列

    分析性质后发现需要一个能删,能合的数据结构,左偏树即可,用堆维护

    #include <cstdio>
    #include <iostream>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #define R(a,b,c) for(register int a = (b); (a) <= (c); ++(a))
    #define nR(a,b,c) for(register int a = (b); (a) >= (c); --(a))
    #define Fill(a,b) memset(a, b, sizeof(a))
    #define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))
    #define ll long long
    #define u32 unsigned int
    #define u64 unsigned long long
     
    #define ON_DEBUGG
     
    #ifdef ON_DEBUGG
     
    #define D_e_Line printf("
    ----------
    ")
    #define D_e(x) cout << (#x) << " : " << x << endl
    #define Pause() system("pause")
    #define FileOpen() freopen("in.txt", "r", stdin)
    #define FileSave() freopen("out.txt", "w", stdout)
    #include <ctime>
    #define TIME() fprintf(stderr, "
    time: %.3fms
    ", clock() * 1000.0 / CLOCKS_PER_SEC)
    
    #else
     
    #define D_e_Line ;
    #define D_e(x) ;
    #define Pause() ;
    #define FileOpen() ;
    #define FileSave() ;
    #define TIME() ;
    //char buf[1 << 21], *p1 = buf, *p2 = buf;
    //#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
     
    #endif
     
    using namespace std;
    struct ios{
        template<typename ATP>inline ios& operator >> (ATP &x){
            x = 0; int f = 1; char ch;
            for(ch = getchar(); ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') f = -1;
            while(ch >= '0' && ch <= '9') x = x * 10 + (ch ^ '0'), ch = getchar();
            x *= f;
            return *this;
        }
    }io;
     
    template<typename ATP>inline ATP Max(ATP a, ATP b){
        return a > b ? a : b;
    }
    template<typename ATP>inline ATP Min(ATP a, ATP b){
        return a < b ? a : b;
    }
    template<typename ATP>inline ATP Abs(ATP a){
        return a < 0 ? -a : a;
    }
    
    const int N = 1e6 + 7;
    
    struct Node {
    	int rt, l, r, siz;
    	long long val;
    } sta[N];
    int top;
    int ch[N][2], dis[N];
    
    int a[N];
    
    #define ls ch[x][0]
    #define rs ch[x][1]
    inline int Merge(int x, int y) {
    	if(!x || !y) return x | y;
    	if(a[x] < a[y]) Swap(x, y);
    	rs = Merge(rs, y);
    	if(dis[rs] > dis[ls]) Swap(ls, rs);
    	dis[x] = dis[rs] + 1;
    	return x;
    }
    
    inline int Del(int x) {
    	return Merge(ls, rs);
    }
    
    int main() {
    	int n;
    	io >> n;
    	dis[0] = -1;
    	R(i,1,n){
    		io >> a[i];
    		a[i] -= i;
    	}
    	R(i,1,n){
    		sta[++top] = (Node){ i, i, i, 1, a[i]};
    		while(top != 1 && sta[top - 1].val > sta[top].val){
    		 	--top;
    		 	sta[top].rt = Merge(sta[top].rt, sta[top + 1].rt);
    		 	sta[top].siz += sta[top + 1].siz;
    		 	sta[top].r = sta[top + 1].r;
    		 	while(sta[top].siz > (sta[top].r - sta[top].l + 2) / 2){
    		 		--sta[top].siz;
    		 		sta[top].rt = Del(sta[top].rt);
    			 }
    			 sta[top].val = a[sta[top].rt];
    		}
    	}
    	int tot = 1;
    	long long ans = 0;
    	R(i,1,n){
    		if(sta[tot].r < i) ++tot;
    		ans += Abs(sta[tot].val - a[i]);
    	}
    	printf("%lld
    ", ans);
    	tot = 1;
    	R(i,1,n){
    		if(sta[tot].r < i) ++tot;
    		printf("%lld ", sta[tot].val + i);
    	}
    	return 0;
    }
    

    咕咕咕

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  • 原文地址:https://www.cnblogs.com/bingoyes/p/11652834.html
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