1. 程式人生 > >【補充上一篇】網路流問題——最大帶權閉合路徑 2018 ACM-ICPC 中國大學生程式設計競賽線上賽 F. Clever King

【補充上一篇】網路流問題——最大帶權閉合路徑 2018 ACM-ICPC 中國大學生程式設計競賽線上賽 F. Clever King

今天把吉林大學的模板也試了一下,它使用的是所謂的Dinic演算法,比普通的BFS不知道快(高)到哪裡去了
程式碼:

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
#define N 400+5 
#define typec long long // type of cost
const typec inf = 0x3f3f3f3f; // max of cost
struct edge {
    int x, y, nxt;
    typec c;
} bf[5000
]; int ne, head[N], cur[N], ps[N], dep[N]; void addedge(int x, int y, typec c) { // add an arc(x -> y, c); vertex: 0 ~ n-1; bf[ne].x = x; bf[ne].y = y; bf[ne].c = c; bf[ne].nxt = head[x]; head[x] = ne++; bf[ne].x = y; bf[ne].y = x; bf[ne].c = 0; bf[ne].nxt = head[y]; head[y] = ne++; } typec flow(int
n, int s, int t) { typec tr, res = 0; int i, j, k, f, r, top; while (1) { memset(dep, -1, n * sizeof(int)); for (f = dep[ps[0] = s] = 0, r = 1; f != r; ) for (i = ps[f++], j = head[i]; j; j = bf[j].nxt) { if (bf[j].c && -1 == dep[k = bf[j].y]) { dep[k] = dep[i] + 1
; ps[r++] = k; if (k == t) { f = r; break; } } } if (-1 == dep[t]) break; memcpy(cur, head, n * sizeof(int)); for (i = s, top = 0; ; ) { if (i == t) { for (k = 0, tr = inf; k < top; ++k) if (bf[ps[k]].c < tr) tr = bf[ps[f = k]].c; for (k = 0; k < top; ++k) bf[ps[k]].c -= tr, bf[ps[k]^1].c += tr; res += tr; i = bf[ps[top = f]].x; } for (j=cur[i]; cur[i]; j = cur[i] = bf[cur[i]].nxt) if (bf[j].c && dep[i]+1 == dep[bf[j].y]) break; if (cur[i]) { ps[top++] = cur[i]; i = bf[cur[i]].y; } else { if (0 == top) break; dep[i] = -1; i = bf[ps[--top]].x; } } } return res; } int main() { int T; int n, m, a, b, t; long long sum; long long mincut; cin >> T; while (T--) { sum = 0; ne = 2; memset(head,0,sizeof(head)); memset(bf,0,sizeof(bf)); cin >> n >> m; for (int i = 1; i <= n; ++i) { cin >> t; addedge(0,i,t); sum += t; } for (int i = 1; i <= m; ++i) { cin >> t; addedge(i+n,n+m+1,t); } for (int i = 1; i <= n; ++i) { cin >> a >> b; for (int j = 0; j<a; ++j) { cin >> t; addedge(i,t+n,inf); } for (int j = 0; j<b; ++j) { cin >> t; addedge(i,t,inf); } } mincut = sum - flow(n+m+2,0,n+m+1); cout << mincut << endl; } return 0; }

最後是一個比較
吉林大學模板:
這裡寫圖片描述
浙江大學模板
這裡寫圖片描述
劉汝佳書上模板:
這裡寫圖片描述