标签:直接 space mat printf typedef efi hdu mon ini
题意:。
析:最小费用流,建立一个超级源点 s 和汇点 t,然后从s 向每个地区边一条容量是 b,费用是a,从每个地区从 t 连一条容量为 d,费用为 -c的边,注意是 -c,然后每个地区有路的就直接连上就好,然后在增广的时候,增广的时候到正数的时候就停止。
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<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1000 + 10; const int maxm = 1e5 + 10; const int mod = 30007; 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, cost; }; struct Mcmf{ int n, m, s, t; vector<Edge> edges; vector<int> G[maxn]; bool inq[maxn]; int d[maxn]; int p[maxn]; int a[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 cap, int cost){ edges.pb((Edge){from, to, cap, 0, cost}); edges.pb((Edge){to, from, 0, 0, -cost}); m = edges.sz; G[from].pb(m-2); G[to].pb(m-1); } bool bellman(int &flow, int &cost){ ms(inq, 0); for(int i = 0; i <= n; ++i) d[i] = -INF; d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF; queue<int> q; q.push(s); while(!q.empty()){ int u = q.front(); q.pop(); inq[u] = 0; for(int i = 0; i < G[u].sz; ++i){ Edge &e = edges[G[u][i]]; if(e.cap > e.flow && d[e.to] < d[u] + e.cost){ d[e.to] = d[u] + e.cost; p[e.to] = G[u][i]; a[e.to] = min(a[u], e.cap - e.flow); if(!inq[e.to]){ q.push(e.to); inq[e.to] = 1; } } } } // printf("%d %d %d\n", cost, d[t], a[t]); if(d[t] < 0) return false; int u = t; while(u != s){ edges[p[u]].flow += a[t]; edges[p[u]^1].flow -= a[t]; u = edges[p[u]].from; } cost += d[t] * a[t]; flow += a[t]; return true; } int mincost(int s, int t){ int flow = 0, cost = 0; this->s = s; this-> t = t; while(bellman(flow, cost)); return cost; } }; Mcmf mcmf; int main(){ while(scanf("%d %d", &n, &m) == 2){ int s = 0, t = n + 1; mcmf.init(t + 2); for(int i = 1; i <= n; ++i){ int a, b, c, d; scanf("%d %d %d %d", &a, &b, &c, &d); mcmf.addEdge(s, i, b, -a); mcmf.addEdge(i, t, d, c); } while(m--){ int u, v, c; scanf("%d %d %d", &u, &v, &c); if(u == v) continue; mcmf.addEdge(u, v, INF, -c); mcmf.addEdge(v, u, INF, -c); } printf("%d\n", mcmf.mincost(s, t)); } return 0; }
标签:直接 space mat printf typedef efi hdu mon ini
原文地址:http://www.cnblogs.com/dwtfukgv/p/7637107.html