NC51281. Cable TV Network
描述
输入描述
Write a program that reads several data sets from the standard input and computes the safety factor for the cable networks encoded by the data sets. Each data set starts with two integers: ,the number of relays in the net, and m, the number of cables in the net. Follow m data pairs (u,v), u < v, where u and v are relay identifiers (integers in the range 0..n-1). The pair (u,v) designates the cable that interconnects the relays u and v. The pairs may occur in any order.Except the (u,v) pairs, which do not contain white spaces, white spaces can occur freely in input. Input data terminate with an end of file and are correct.
输出描述
For each data set, the program prints on the standard output, from the beginning of a line, the safety factor of the encoded net.
示例1
输入:
0 0 1 0 3 3 (0,1) (0,2) (1,2) 2 0 5 7 (0,1) (0,2) (1,3) (1,2) (1,4) (2,3) (3,4)
输出:
0 1 3 0 2
说明:
The first data set encodes an empty network, the second data set corresponds to a network with a single relay, and the following three data sets encode the nets shown in figure 1.C++14(g++5.4) 解法, 执行用时: 28ms, 内存消耗: 376K, 提交时间: 2020-08-01 22:12:26
#include <queue> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int N = 56, M = 20006, INF = 0x3f3f3f3f; int n, m, s, t; int a[N*N], b[N*N], d[N<<1]; int Head[N<<1], Edge[M], Leng[M], Next[M], tot; inline void add(int x, int y, int z) { Edge[++tot] = y; Leng[tot] = z; Next[tot] = Head[x]; Head[x] = tot; Edge[++tot] = x; Leng[tot] = 0; Next[tot] = Head[y]; Head[y] = tot; } inline bool bfs() { memset(d, 0, sizeof(d)); queue<int> q; q.push(s); d[s] = 1; while (q.size()) { int x = q.front(); q.pop(); for (int i = Head[x]; i; i = Next[i]) { int y = Edge[i], z = Leng[i]; if (z && !d[y]) { q.push(y); d[y] = d[x] + 1; if (y == t) return 1; } } } return 0; } inline int dinic(int x, int f) { if (x == t) return f; int rest = f; for (int i = Head[x]; i && rest; i = Next[i]) { int y = Edge[i], z = Leng[i]; if (z && d[y] == d[x] + 1) { int k = dinic(y, min(rest, z)); if (!k) d[y] = 0; Leng[i] -= k; Leng[i^1] += k; rest -= k; } } return f - rest; } inline void Cable_TV_Network() { for (int i = 0; i < m; i++) { char str[20]; scanf("%s", str); a[i] = b[i] = 0; int j; for (j = 1; str[j] != ','; j++) a[i] = a[i] * 10 + str[j] - '0'; for (j++; str[j] != ')'; j++) b[i] = b[i] * 10 + str[j] - '0'; } int ans = INF; for (s = 0; s < n; s++) for (t = 0; t < n; t++) if (s != t) { memset(Head, 0, sizeof(Head)); tot = 1; int maxf = 0; for (int i = 0; i < n; i++) if (i == s || i == t) add(i, i + n, INF); else add(i, i + n, 1); for (int i = 0; i < m; i++) { add(a[i] + n, b[i], INF); add(b[i] + n, a[i], INF); } while (bfs()) { int num; while ((num = dinic(s, INF))) maxf += num; } ans = min(ans, maxf); } if (n <= 1 || ans == INF) ans = n; cout << ans << endl; } int main() { while (cin >> n >> m) Cable_TV_Network(); return 0; }
C++ 解法, 执行用时: 16ms, 内存消耗: 1064K, 提交时间: 2022-04-29 19:18:39
#include<stdio.h> #include<string.h> #include<queue> #include<algorithm> #define inf 0x3f3f3f3f using namespace std; int n,m; int mp[300][300],pre[300],flow[300][300],p[300],a[300]; int EK(int s,int t) { int sum=0; int m=2*s+2; queue<int>q; memset(flow,0,sizeof(flow)); for(;;) { memset(a,0,sizeof(a)); a[s]=inf; q.push(s); while(!q.empty()) { int u=q.front(); q.pop(); for(int i=0;i<m;i++)if(!a[i]&&mp[u][i]>flow[u][i])p[i]=u,q.push(i),a[i]=a[u]<mp[u][i]-flow[u][i]?a[u]:mp[u][i]-flow[u][i]; } if(!a[t])break; for(int i=t; i!=s; i=p[i])flow[p[i]][i]+=a[t],flow[i][p[i]]-=a[t]; sum+=a[t]; } return sum; } int main(){ int a,b; while(scanf("%d%d",&n,&m)!=EOF){ memset(mp,0,sizeof(mp)); memset(p,0,sizeof(p)); for(int i=0;i<n;i++)mp[i][i+n]=1; while(m--){ scanf(" (%d,%d)",&a,&b); mp[a+n][b]=inf; mp[b+n][a]=inf; } int ans=inf; for(int i=1;i<n;i++)ans=min(ans,EK(n,i)); if(ans==inf)ans=n; printf("%d\n",ans); } return 0; }