标签:type 最大的 mat 1.0 back cst tor init stack
题意:给定n个带权点m条无向带权边,选一个子图,则这个子图的权值为 边权和-点权和,求一个最大的权值。
析:把每条边都看成是一个新点,然后建图,就是一个裸的最大闭合子图。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #include <assert.h> #include <bitset> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a, b, sizeof a) #define sz size() #define pu push_up #define pd push_down #define cl clear() #define all 1,n,1 #define FOR(i,x,n) for(int i = (x); i < (n); ++i) #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<LL, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e17; const double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 55000 + 50; const int maxm = 1e6 + 5; const int mod = 10007; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, -1, 0, 1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c) { return r >= 0 && r < n && c >= 0 && c < m; } struct Edge{ int from, to, cap, flow; }; struct Dinic{ int n, m, s, t; vector<Edge> edges; vector<int> G[maxn]; bool vis[maxn]; int d[maxn]; int cur[maxn]; void init(int n){ this->n = n; for(int i = 0; i < n; ++i) G[i].cl; edges.cl; } void addEdge(int from, int to, int c){ edges.pb((Edge){from, to, c, 0}); edges.pb((Edge){to, from, 0, 0}); m = edges.sz; G[from].pb(m - 2); G[to].pb(m - 1); } bool bfs(){ ms(vis, 0); vis[s] = 1; d[s] = 0; queue<int> q; q.push(s); while(!q.empty()){ int u = q.front(); q.pop(); for(int i = 0; i < G[u].sz; ++i){ Edge &e = edges[G[u][i]]; if(!vis[e.to] && e.cap > e.flow){ d[e.to] = d[u] + 1; vis[e.to] = 1; q.push(e.to); } } } return vis[t]; } int dfs(int u, int a){ if(u == t || a == 0) return a; int flow = 0, f; for(int &i = cur[u]; i < G[u].sz; ++i){ Edge &e = edges[G[u][i]]; if(d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0){ e.flow += f; edges[G[u][i]^1].flow -= f; flow += f; a -= f; if(a == 0) break; } } return flow; } int maxflow(int s, int t){ this->s = s; this->t = t; int flow = 0; while(bfs()){ ms(cur, 0); flow += dfs(s, INF); } return flow; } }; Dinic dinic; int main(){ while(scanf("%d %d", &n, &m) == 2){ int s = 0, t = n + m + 1; dinic.init(t + 5); for(int i = 1; i <= n; ++i){ int c; scanf("%d", &c); dinic.addEdge(i, t, c); } int sum = 0; for(int i = 1; i <= m; ++i){ int u, v, c; scanf("%d %d %d", &u, &v, &c); dinic.addEdge(n + i, u, INF); dinic.addEdge(n + i, v, INF); dinic.addEdge(s, n + i, c); sum += c; } printf("%d\n", sum - dinic.maxflow(s, t)); } return 0; }
HDU 3897 Base Station (网络流,最大闭合子图)
标签:type 最大的 mat 1.0 back cst tor init stack
原文地址:http://www.cnblogs.com/dwtfukgv/p/7676701.html