费用流问题,每个样本选一次,就连一条capacity为1,权为给定的值,因为可以重复走,再连capacity为无穷,权为0的边,再一次连接给定的出点和汇点即可
#include<bits/stdc++.h> using namespace std; #define lowbit(x) ((x)&(-x)) typedef long long LL; const int maxm = 3e3+5; const int INF = 0x3f3f3f3f; struct edge{ int u, v, cap, flow, cost, nex; } edges[maxm]; int head[maxm], cur[maxm], cnt, fa[1024], d[1024], p, q; bool inq[1024]; void init() { memset(head, -1, sizeof(head)); } void add(int u, int v, int cap, int cost) { edges[cnt] = edge{u, v, cap, 0, cost, head[u]}; head[u] = cnt++; } void addedge(int u, int v, int cap, int cost) { add(u, v, cap, cost), add(v, u, 0, -cost); } bool spfa(int s, int t, int &flow, LL &cost) { for(int i = 0; i <= p*q+3; ++i) d[i] = INF; //init() memset(inq, false, sizeof(inq)); d[s] = 0, inq[s] = true; fa[s] = -1, cur[s] = INF; queue<int> q; q.push(s); while(!q.empty()) { int u = q.front(); q.pop(); inq[u] = false; for(int i = head[u]; i != -1; i = edges[i].nex) { edge& now = edges[i]; int v = now.v; if(now.cap > now.flow && d[v] > d[u] + now.cost) { d[v] = d[u] + now.cost; fa[v] = i; cur[v] = min(cur[u], now.cap - now.flow); if(!inq[v]) {q.push(v); inq[v] = true;} } } } if(d[t] == INF) return false; flow += cur[t]; cost += 1LL*d[t]*cur[t]; for(int u = t; u != s; u = edges[fa[u]].u) { edges[fa[u]].flow += cur[t]; edges[fa[u]^1].flow -= cur[t]; } return true; } int MincostMaxflow(int s, int t, LL &cost) { cost = 0; int flow = 0; while(spfa(s, t, flow, cost)); return flow; } void run_case() { init(); int a, b, val; cin >> a >> b >> p >> q; p++, q++; for(int i = 1; i <= p; ++i) for(int j = 1; j < q; ++j) { cin >> val; addedge((i-1)*q+j, (i-1)*q+j+1, 1, -val); addedge((i-1)*q+j, (i-1)*q+j+1, INF, 0); } for(int i = 1; i <= q; ++i) for(int j = 1; j < p; ++j) { cin >> val; addedge((j-1)*q+i, j*q+i, 1, -val); addedge((j-1)*q+i, j*q+i, INF, 0); } int s = 0, t = p*q+2; for(int i = 0; i < a; ++i) { int k, x, y; cin >> k >> x >> y; x++, y++; addedge(s, (x-1)*q+y, k, 0); } for(int i = 0; i < b; ++i) { int k, x, y; cin >> k >> x >> y; x++, y++; addedge((x-1)*q+y, t, k, 0); } LL cost = 0; MincostMaxflow(s, t, cost); cout << -cost; } int main() { ios::sync_with_stdio(false), cin.tie(0); run_case(); cout.flush(); return 0; }
来源:https://www.cnblogs.com/GRedComeT/p/12299123.html