标签：als https else dinic names double oid blank push
[题目链接]
https://www.lydsy.com/JudgeOnline/problem.php?id=4873
[算法]
注意到题目中的限制条件可表述为 : 若选择区间[L , R] , 则必须选择区间[L + 1 , R]和[L , R - 1] ， 这种依赖关系可以让我们联想到用最大权闭合子图解题
将每种代号建一个点 ， 每个区间同样建一个点
首先将每个形如[i , i]的区间向其代号连边
然后将每个区间[L , R]所代表的点向[L + 1 , R]和[L , R - 1]连边
注意我们需要减去代价mx ^ 2 + cx
那么我们将每个形如[i , i]的区间所代表点的点权减去其代号 ， 将每种代号i所代表点的点权减去m * i ^ 2
时间复杂度 : O(Dinic(N ^ 2 , N ^ 2))
[代码]
#include<bits/stdc++.h> using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; const int N = 110; const int inf = 2e9; struct edge { int to , w , nxt; } e[N * N * 10]; int n , m , cnt , mx , S , T , tot; int d[N][N] , a[N * 10] , point[N][N] , head[N * N * 10] , dep[N * N * 10]; template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); } template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); } template <typename T> inline void read(T &x) { T f = 1; x = 0; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f; for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘; x *= f; } inline void addedge(int u , int v , int w) { ++tot; e[tot] = (edge){v , w , head[u]}; head[u] = tot; ++tot; e[tot] = (edge){u , 0 , head[v]}; head[v] = tot; } inline bool bfs() { queue< int > q; for (int i = 1; i <= T; ++i) dep[i] = -1; q.push(S); dep[S] = 1; while (!q.empty()) { int cur = q.front(); q.pop(); for (int i = head[cur]; i; i = e[i].nxt) { int v = e[i].to , w = e[i].w; if (w > 0 && dep[v] == -1) { dep[v] = dep[cur] + 1; q.push(v); if (v == T) return true; } } } return false; } inline int dinic(int u , int flow) { int k , rest = flow; if (u == T) return flow; for (int i = head[u]; i && rest; i = e[i].nxt) { int v = e[i].to , w = e[i].w; if (w > 0 && dep[v] == dep[u] + 1) { k = dinic(v , min(rest , w)); e[i].w -= k; e[i ^ 1].w += k; if (!k) dep[v] = 0; rest -= k; } } return flow - rest; } int main() { read(n); read(m); tot = 1; for (int i = 1; i <= n; ++i) { read(a[i]); mx = max(mx , a[i]); } cnt = mx; for (int i = 1; i <= n; ++i) { for (int j = i; j <= n; ++j) { read(d[i][j]); point[i][j] = ++cnt; } } S = cnt + 1 , T = S + 1; int ans = 0; for (int i = 1; i <= n; ++i) d[i][i] -= a[i]; for (int i = 1; i <= mx; ++i) addedge(i , T , m * i * i); for (int i = 1; i <= n; i++) addedge(point[i][i] , a[i] , inf); for (int i = 1; i <= n; ++i) { for (int j = i; j <= n; ++j) { if (i <= j - 1) addedge(point[i][j] , point[i][j - 1] , inf); if (i + 1 <= j) addedge(point[i][j] , point[i + 1][j] , inf); if (d[i][j] >= 0) { ans += d[i][j]; addedge(S , point[i][j] , d[i][j]); } else addedge(point[i][j] , T , -d[i][j]); } } while (bfs()) { while (int flow = dinic(S , inf)) ans -= flow; } printf("%d\n" , ans); return 0; }
标签：als https else dinic names double oid blank push
原文地址：https://www.cnblogs.com/evenbao/p/10540066.html
