luogu P3358 最长k可重区间集问题
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网络流建图好难,这题居然是网络流(雾,一般分析来说,有限制的情况最大流情况可以拆点通过capacity来限制,比如只使用一次,把一个点拆成入点出点,capacity为1即可,这题是限制最大k重复,可以联想到最大流问题,设源点汇点,限制的k就是其最大的capacity,其最大流一定<=k,跑出来一定满足条件,但如何计算长度呢,就使用费用流吧,最大费用流就把边权取反即可,我们不知道输入的数据范围,只知道个数,就离散化一下,每个区间只能选一次,就对离散化后的l对r连一条capacity为1,权为-len的边,源点对1连一条capacity为k的边,n对汇点连一条capacity为k的边,因为判断是否重复是从头到尾,那么源点汇点就分别连接头尾,但是这样连图不一定保证每个点都考虑进去了,可能会出现图不连通的情况,题目是开区间,那么(i,i+1)和(i+1,i+2)一定是没有交点的,就对i与i+1连一条capacity为无穷,权为0的边,这样就能保证图连通且不会影响到答案,因为如果有断层,其会顺着往下搜索,权为0,capacity为无穷,对答案无影响
#include<bits/stdc++.h> using namespace std; #define lowbit(x) ((x)&(-x)) typedef long long LL; const int maxm = 5e3+5; const int INF = 0x3f3f3f3f; struct edge{ int u, v, cap, flow, cost, nex; } edges[maxm]; struct Points{ int l, r, len; } point[505]; int head[maxm], cur[maxm], cnt, fa[1024], d[1024], n, allx[1024]; bool inq[1024]; void init() { memset(head, -1, sizeof(head)); } void add(int u, int v, int cap, int cost) { edges[cnt] = edge{u, v, cap, 0, cost, head[u]}; head[u] = cnt++; } void addedge(int u, int v, int cap, int cost) { add(u, v, cap, cost), add(v, u, 0, -cost); } bool spfa(int s, int t, int &flow, LL &cost) { for(int i = 0; i <= ((n<<1)|1); ++i) d[i] = INF; //init() memset(inq, false, sizeof(inq)); d[s] = 0, inq[s] = true; fa[s] = -1, cur[s] = INF; queue<int> q; q.push(s); while(!q.empty()) { int u = q.front(); q.pop(); inq[u] = false; for(int i = head[u]; i != -1; i = edges[i].nex) { edge& now = edges[i]; int v = now.v; if(now.cap > now.flow && d[v] > d[u] + now.cost) { d[v] = d[u] + now.cost; fa[v] = i; cur[v] = min(cur[u], now.cap - now.flow); if(!inq[v]) {q.push(v); inq[v] = true;} } } } if(d[t] == INF) return false; flow += cur[t]; cost += 1LL*d[t]*cur[t]; for(int u = t; u != s; u = edges[fa[u]].u) { edges[fa[u]].flow += cur[t]; edges[fa[u]^1].flow -= cur[t]; } return true; } int MincostMaxflow(int s, int t, LL &cost) { cost = 0; int flow = 0; while(spfa(s, t, flow, cost)); return flow; } void run_case() { init(); int l, r, k, xcnt = 0; cin >> n >> k; for(int i = 1; i <= n; ++i) { cin >> l >> r; allx[++xcnt] = l, allx[++xcnt] = r, point[i] = Points{l, r, r-l}; } sort(allx+1,allx+1+xcnt); int len = unique(allx+1,allx+1+xcnt)-allx; for(int i = 1; i <= n; ++i) { point[i].l = lower_bound(allx+1,allx+len,point[i].l)-allx; point[i].r = lower_bound(allx+1,allx+len,point[i].r)-allx; } for(int i = 1; i < len-1; ++i) addedge(i, i+1, INF, 0); int s = 0, t = len; for(int i = 1; i <= n; ++i) { addedge(point[i].l, point[i].r, 1, -point[i].len); } addedge(s, 1, k, 0), addedge(len-1, t, k, 0); LL cost = 0; MincostMaxflow(s, t, cost); cout << -cost; } int main() { ios::sync_with_stdio(false), cin.tie(0); run_case(); cout.flush(); return 0; }
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