POJ2516

题目链接：http://poj.org/problem?id=2516

解题思路：

　　最小费用最大流，这个没什么疑问。但此题小难点在于读题，大难点在于建图。

　　首先，供应量小于需求量的时候直接输出“-1”。

　　供大于或等于求的情况，一开始我将每个供应商和每个购买人都拆成K个点，将所有供应商的点和超级源点相连，流量限制为库存量，费用为0；将所有购买人的点和超级汇点相连，流量限制为购买量，费用为0；而购买方和供应方之间的连线的流量限制则为inf，费用如题目中给出的。但是这种建图方式一直T......一度绝望到以为我的模板有问题，修修补补了半天，还是T......上网查了题解，发现大家的建图方式跟我不一样：大家都是把根据K种商品，建K个图逐一求最小费用最大流，答案就是K个图的最小费用之和。稍微改了一下，AC.......醉........至于为什么........我也不太懂，容我明天去问问师兄。

AC代码：

  1 #include <cstdio>
  2 #include <vector>
  3 #include <queue>
  4 #include <cstring>
  5 using namespace std;
  6
  7 const int MAXN = 5000;
  8 const int MAXM = 100000;
  9 const int INF = 0x3f3f3f3f;
 10 struct rec {
 11     int from, to, cost, cap;
 12 }r[55][2550];
 13 struct Edge {
 14     int to, next, cap, flow, cost;
 15 }edge[MAXM];
 16 int head[MAXN], tol;
 17 int pre[MAXN], dis[MAXN];
 18 bool vis[MAXN];
 19 int N;
 20 void init(int n) {
 21     N = n;
 22     tol = 0;
 23     memset(head, -1, sizeof(head));
 24 }
 25 void addedge(int u, int v, int cap, int cost) {
 26     edge[tol].to = v;
 27     edge[tol].cap = cap;
 28     edge[tol].cost = cost;
 29     edge[tol].flow = 0;
 30     edge[tol].next = head[u];
 31     head[u] = tol++;
 32     edge[tol].to = u;
 33     edge[tol].cap = 0;
 34     edge[tol].cost = -cost;
 35     edge[tol].flow = 0;
 36     edge[tol].next = head[v];
 37     head[v] = tol++;
 38 }
 39 bool spfa(int s, int t) {
 40     queue<int>q;
 41     for (int i = 0; i < N; i++) {
 42         dis[i] = INF;
 43         vis[i] = false;
 44         pre[i] = -1;
 45     }
 46     dis[s] = 0;
 47     vis[s] = true;
 48     q.push(s);
 49     while (!q.empty()) {
 50         int u = q.front();
 51         q.pop();
 52         vis[u] = false;
 53         for (int i = head[u]; i != -1; i = edge[i].next) {
 54             int v = edge[i].to;
 55             if (edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost){
 56                 dis[v] = dis[u] + edge[i].cost;
 57                 pre[v] = i;
 58                 if (!vis[v]){
 59                     vis[v] = true;
 60                     q.push(v);
 61                 }
 62             }
 63         }
 64     }
 65     if (pre[t] == -1)    return false;
 66     else    return true;
 67 }
 68 int minCostMaxflow(int s, int t, int &cost){
 69     int flow = 0;
 70     cost = 0;
 71     while (spfa(s, t)) {
 72         int Min = INF;
 73         for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to]) {
 74             if (Min > edge[i].cap - edge[i].flow)
 75                 Min = edge[i].cap - edge[i].flow;
 76         }
 77         for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to]) {
 78             edge[i].flow += Min;
 79             edge[i ^ 1].flow -= Min;
 80             cost += edge[i].cost * Min;
 81         }
 82         flow += Min;
 83     }
 84     return flow;
 85 }
 86 int sup[55][55], buy[55][55];
 87 int main() {
 88     int N, M, K;
 89     int need[53], have[53];
 90     while (scanf("%d%d%d", &N, &M, &K) == 3 && (N || M || K)) {
 91         memset(need, 0, sizeof(need));
 92         memset(have, 0, sizeof(have));
 93         int c;
 94         for (int i = 1; i <= N; i++) {
 95             for (int j = 1; j <= K; j++) {
 96                 scanf("%d", &c);
 97                 need[j] += c;
 98                 buy[j][i] = c;
 99             }
100         }
101         for (int i = 1; i <= M; i++) {
102             for (int j = 1; j <= K; j++) {
103                 scanf("%d", &c);
104                 have[j] += c;
105                 sup[j][i] = c;
106             }
107         }
108         bool yes = true;
109         for (int j = 1; j <= K; j++) {
110             if (have[j]<need[j]) yes = false;
111         }
112
113         int ans = 0;
114         for (int k = 1; k <= K; k++) {
115             if (yes)
116                 init(N + M + 2);
117             for (int i = 1; i <= N; i++) {
118                 for (int j = 1; j <= M; j++) {
119                     scanf("%d", &c);
120                     if (yes)
121                         addedge(j, M + i, INF, c);
122                 }
123             }
124             if (yes) {
125                 for (int i = 1; i <= M; i++)
126                     addedge(0, i, sup[k][i], 0);
127                 for (int i = 1; i <= N; i++)
128                     addedge(M + i, N + M + 1, buy[k][i], 0);
129                 int cost;
130                 minCostMaxflow(0, N + M + 1, cost);
131                 ans += cost;
132             }
133         }
134         if (!yes)
135             printf("-1\n");
136         else
137             printf("%d\n", ans);
138     }
139     return 0;
140 }