bzoj3638
费用流+线段树
看见这个题我们马上就能想到费用流,设立源汇,分别向每个点连接容量为1费用为0的边,然后相邻的点之间连边,费用为点权,跑费用流就行了,但是很明显这样会超时,那么我们要优化一下,我们观察费用流的过程,发现对于点与点之间的边,每次取一段区间相当于把正向边改为反向边,费用变负,于是我们可以用线段树来模拟这个过程,像费用流一样贪心地选取区间的最大子段和,然后取反,每次取k次,然后恢复。这样就好了
但是写的时候有很多问题,比如如何返回一个区间?结构体!参考了popoqqq大神的代码,发现我们可以通过重载小于号直接对结构体取max,这样就十分好写了
然后这道题有点卡常,一定要在重载的时候把传入参数变成const+引用,这样在cf上快了200ms
这就是传说中的五倍经验吗
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<bits/stdc++.h> using namespace std; const int N = 100010; namespace IO { const int Maxlen = N * 50; char buf[Maxlen], *C = buf; int Len; inline void read_in() { Len = fread(C, 1, Maxlen, stdin); buf[Len] = '\0'; } inline void fread(int &x) { x = 0; int f = 1; while (*C < '0' || '9' < *C) { if(*C == '-') f = -1; ++C; } while ('0' <= *C && *C <= '9') x = (x << 1) + (x << 3) + *C - '0', ++C; x *= f; } inline void read(int &x) { x = 0; int f = 1; char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { x = (x << 1) + (x << 3) + c - '0'; c = getchar(); } x *= f; } inline void read(long long &x) { x = 0; long long f = 1; char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { x = (x << 1ll) + (x << 3ll) + c - '0'; c = getchar(); } x *= f; } } using namespace IO; struct data { int l, r, v; data() {} data(int l, int r, int v) : l(l), r(r), v(v) {} friend bool operator < (const data &a, const data &b) { return a.v < b.v; } friend data operator + (const data &a, const data &b) { return data(a.l, b.r, a.v + b.v); } }; struct node { data lmax, rmax, mx, mn, lmin, rmin, sum; int tag; node() {} node(int x, int v) { lmax = rmax = mx = mn = lmin = rmin = sum = data(x, x, v); } friend node operator + (const node &a, const node &b) { node c; if(a.tag == -1) return b; if(b.tag == -1) return a; c.tag = 0; c.sum = a.sum + b.sum; c.lmax = max(a.lmax, a.sum + b.lmax); c.lmin = min(a.lmin, a.sum + b.lmin); c.rmax = max(b.rmax, a.rmax + b.sum); c.rmin = min(b.rmin, a.rmin + b.sum); c.mx = max(max(a.mx, b.mx), a.rmax + b.lmax); c.mn = min(min(a.mn, b.mn), a.rmin + b.lmin); return c; } } tree[N << 2], st[21]; int n, q; int a[N]; void paint(node &o) { swap(o.lmax, o.lmin); swap(o.rmax, o.rmin); swap(o.mx, o.mn); o.sum.v *= -1; o.lmax.v *= -1; o.lmin.v *= -1; o.rmax.v *= -1; o.rmin.v *= -1; o.mx.v *= -1; o.mn.v *= -1; o.tag ^= 1; } void pushdown(int x) { if(tree[x].tag <= 0) return; paint(tree[x << 1]); paint(tree[x << 1 | 1]); tree[x].tag ^= 1; } void build(int l, int r, int x) { if(l == r) { tree[x] = node(l, a[l]); return; } int mid = (l + r) >> 1; build(l, mid, x << 1); build(mid + 1, r, x << 1 | 1); tree[x] = tree[x << 1] + tree[x << 1 | 1]; } node query(int l, int r, int x, int a, int b) { if(l > b || r < a) return tree[0]; if(l >= a && r <= b) return tree[x]; pushdown(x); int mid = (l + r) >> 1; return (query(l, mid, x << 1, a, b)) + (query(mid + 1, r, x << 1 | 1, a, b)); } void reverse(int l, int r, int x, int a, int b) { if(l > b || r < a) return; if(l >= a && r <= b) { paint(tree[x]); return; } pushdown(x); int mid = (l + r) >> 1; reverse(l, mid, x << 1, a, b); reverse(mid + 1, r, x << 1 | 1, a, b); tree[x] = tree[x << 1] + tree[x << 1 | 1]; } void update(int l, int r, int x, int pos, int v) { if(l == r) { tree[x] = node(l, v); return; } pushdown(x); int mid = (l + r) >> 1; if(pos <= mid) update(l, mid, x << 1, pos, v); else update(mid + 1, r, x << 1 | 1, pos, v); tree[x] = tree[x << 1] + tree[x << 1 | 1]; } int main() { read_in(); fread(n); for(int i = 1; i <= n; ++i) fread(a[i]); tree[0].tag = -1; build(1, n, 1); fread(q); while(q--) { int opt, l, r, v; fread(opt); if(opt == 0) { fread(l); fread(v); update(1, n, 1, l, v); } if(opt == 1) { fread(l); fread(r); fread(v); int sum = 0, top = 0; while(v--) { node ans = query(1, n, 1, l, r); if(ans.mx.v <= 0) break; reverse(1, n, 1, ans.mx.l, ans.mx.r); node tmp = query(1, n, 1, ans.mx.l, ans.mx.r); st[++top] = ans; sum += ans.mx.v; } printf("%d\n", sum); for(int i = top; i; --i) reverse(1, n, 1, st[i].mx.l, st[i].mx.r); } } return 0; }