最小费用最大流

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本文介绍了一个基于最小费用最大流算法解决的题目“Kaka的矩阵旅行”。通过构建特殊的图结构，采用连续最短路径增广算法求解，实现从起点到终点的路径寻找，并计算最小费用。

/* ID: linjd821 LANG: C++ TASK: http://acm.pku.edu.cn/JudgeOnline/problem?id=3422 Kaka's Matrix Travels */ /* 算法说明：最小费用最大流输入：链接表图edge[]+head[], 点个数N,源点src, 汇点dest 输出：ans = 最小费用最大流代码参考：谢政《网络算法与复杂性理论》网上应该也有英文版的论文，我这个是连续最短路增广算法（successive shortest path）简称SSP， */ #include <stdio.h> #include <string.h> #include <stdlib.h> #include <math.h> #include <assert.h> #include <ctype.h> #include <map> #include <string> #include <set> #include <bitset> #include <utility> #include <algorithm> #include <vector> #include <stack> #include <queue> #include <iostream> #include <fstream> #include <list> using namespace std; const int MAXN = 5000+10; const int MAXM = 50000; const int INF = 1000000000; struct Edge { int next; int st, ed; int flow; int cost; }edge[MAXM]; int head[MAXN]; int N, K, E, src, dest; int g[55][55]; int d[MAXN], pre[MAXN]; inline void add_edge(int u, int v, int flow, int cost) { edge[E].st = u; edge[E].ed = v; edge[E].flow = flow; edge[E].cost = cost; edge[E].next = head[u]; head[u] = E++; edge[E].st = v; edge[E].ed = u; edge[E].flow = 0; edge[E].cost = -cost; edge[E].next = head[v]; head[v] = E++; } //for every node i add_edge i-->i' flow = 1; cost = g[i][j] //for every path i to j add_edge i'-->j flow = INF; cost = 0; void BuildGraph() { int i, j, u; for(i = 1; i <= N; i++) for(j = 1; j <= N; j++) scanf("%d", &g[i][j]); memset(head, -1, sizeof(head)); E = 0; for(i = 1; i <= N; i++) for(j = 1; j <= N; j++) { add_edge((i-1)*N+j, (i-1)*N+j+N*N, 1, -g[i][j]); if(i != N) { add_edge((i-1)*N+j+N*N, i*N+j, INF, 0); add_edge((i-1)*N+j, i*N+j, INF, 0); } if(j != N) { add_edge((i-1)*N+j+N*N, (i-1)*N+j+1, INF, 0); add_edge((i-1)*N+j, (i-1)*N+j+1, INF, 0); } if(i == N || (i == N-1 && j == N)) { add_edge((i-1)*N+j, 2*N*N, INF, 0); } } add_edge(0, 1, K, 0); src = 0; N = N*N*2; dest = N; } int SPFA() { int que[MAXN][2], rear = 1, front = 0, i, aug = INF; bool used[MAXN]; memset(used, 0, sizeof(used)); for(i = 0; i <= N; i++) d[i] = INF; que[0][0] = src; que[0][1] = INF; d[src] = 0; while(rear != front) { int cur = que[front][0]; int cp = que[front][1]; //printf("cur = %d, cp = %d, d[%d] = %d/n", cur, cp, cur, d[cur]); if(cur == dest) aug = cp; front = (front+1)%MAXN; used[cur] = false; for(i = head[cur]; i != -1; i = edge[i].next) { if(edge[i].flow && d[edge[i].ed] > d[cur]+edge[i].cost) { d[edge[i].ed] = d[cur]+edge[i].cost; pre[edge[i].ed] = i; if(!used[edge[i].ed]) { used[edge[i].ed] = true; que[rear][0] = edge[i].ed; que[rear][1] = min(cp, edge[i].flow); rear = (rear+1)%MAXN; } } } } return aug == INF ? 0 : aug; } int MinCostFlow() { int i, ans = 0, t; while((t = SPFA())) { ans -= d[dest]*t; int ptr = dest; while(ptr != src) { ptr = pre[ptr]; edge[ptr].flow -= t; edge[ptr^1].flow += t; ptr = edge[ptr].st; } } return ans; } int main() { while(scanf("%d %d", &N, &K) != EOF) { BuildGraph(); printf("%d/n", MinCostFlow()); } return 0; }