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：

 #include <cstdio>
 #include <vector>
 #include <queue>
 #include <cstring>
 using namespace std;

 const int MAXN = ;
 const int MAXM = ;
 const int INF = 0x3f3f3f3f;
 struct rec {
     int from, to, cost, cap;
 }r[][];
 struct Edge {
     int to, next, cap, flow, cost;
 }edge[MAXM];
 int head[MAXN], tol;
 int pre[MAXN], dis[MAXN];
 bool vis[MAXN];
 int N;
 void init(int n) {
     N = n;
     tol = ;
     memset(head, -, sizeof(head));
 }
 void addedge(int u, int v, int cap, int cost) {
     edge[tol].to = v;
     edge[tol].cap = cap;
     edge[tol].cost = cost;
     edge[tol].flow = ;
     edge[tol].next = head[u];
     head[u] = tol++;
     edge[tol].to = u;
     edge[tol].cap = ;
     edge[tol].cost = -cost;
     edge[tol].flow = ;
     edge[tol].next = head[v];
     head[v] = tol++;
 }
 bool spfa(int s, int t) {
     queue<int>q;
     for (int i = ; i < N; i++) {
         dis[i] = INF;
         vis[i] = false;
         pre[i] = -;
     }
     dis[s] = ;
     vis[s] = true;
     q.push(s);
     while (!q.empty()) {
         int u = q.front();
         q.pop();
         vis[u] = false;
         for (int i = head[u]; i != -; i = edge[i].next) {
             int v = edge[i].to;
             if (edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost){
                 dis[v] = dis[u] + edge[i].cost;
                 pre[v] = i;
                 if (!vis[v]){
                     vis[v] = true;
                     q.push(v);
                 }
             }
         }
     }
     if (pre[t] == -)    return false;
     else    return true;
 }
 int minCostMaxflow(int s, int t, int &cost){
     int flow = ;
     cost = ;
     while (spfa(s, t)) {
         int Min = INF;
         for (int i = pre[t]; i != -; i = pre[edge[i ^ ].to]) {
             if (Min > edge[i].cap - edge[i].flow)
                 Min = edge[i].cap - edge[i].flow;
         }
         for (int i = pre[t]; i != -; i = pre[edge[i ^ ].to]) {
             edge[i].flow += Min;
             edge[i ^ ].flow -= Min;
             cost += edge[i].cost * Min;
         }
         flow += Min;
     }
     return flow;
 }
 int sup[][], buy[][];
 int main() {
     int N, M, K;
     int need[], have[];
     while (scanf("%d%d%d", &N, &M, &K) ==  && (N || M || K)) {
         memset(need, , sizeof(need));
         memset(have, , sizeof(have));
         int c;
         for (int i = ; i <= N; i++) {
             for (int j = ; j <= K; j++) {
                 scanf("%d", &c);
                 need[j] += c;
                 buy[j][i] = c;
             }
         }
         for (int i = ; i <= M; i++) {
             for (int j = ; j <= K; j++) {
                 scanf("%d", &c);
                 have[j] += c;
                 sup[j][i] = c;
             }
         }
         bool yes = true;
         for (int j = ; j <= K; j++) {
             if (have[j]<need[j]) yes = false;
         }

         int ans = ;
         for (int k = ; k <= K; k++) {
             if (yes)
                 init(N + M + );
             for (int i = ; i <= N; i++) {
                 for (int j = ; j <= M; j++) {
                     scanf("%d", &c);
                     if (yes)
                         addedge(j, M + i, INF, c);
                 }
             }
             if (yes) {
                 for (int i = ; i <= M; i++)
                     addedge(, i, sup[k][i], );
                 for (int i = ; i <= N; i++)
                     addedge(M + i, N + M + , buy[k][i], );
                 int cost;
                 minCostMaxflow(, N + M + , cost);
                 ans += cost;
             }
         }
         if (!yes)
             printf("-1\n");
         else
             printf("%d\n", ans);
     }
     return ;
 }