POJ 1149 解题报告

这道题是网络流问题，难点在于构图。

大神给了清晰易懂的构图解释：http://ycool.com/post/zhhrrm6

之后就转化为很普通的最大流问题，我用的是之前写的dinic代码。

构图的时候存在着这样的问题：如果两个节点之间有多条边，是否合并。两者都是正确的，但合并会直接减少边数，提高运行速率，应该是最好的方法。但是这种方法需要建立一个matrix，而不能再用adjlist。这里偷懒沿用了后者，但是由于这样两个节点之间可能存在多条边，于是预先需要分配多少条边就需要测试了。（在提交了好几次后）这里用了100000。另外，如果程序中有assert。如果fail了，POJ只会返回WA。

thestoryofsnow1149Accepted1324K32MSC++

/* ID: thestor1 LANG: C++ TASK: poj1149 */#include <iostream>#include <fstream>#include <cmath>#include <cstdio>#include <cstring>#include <limits>#include <string>#include <vector>#include <list>#include <set>#include <map>#include <queue>#include <stack>#include <algorithm>#include <cassert>using namespace std;const int MAXN = 100 + 2;const int MAXM = 1000;const int INF = numeric_limits<int>::max();class Edge{public:// edge (`u` -> `v`) with capacity `w`int v, w;// use (pre-allocated) array as linked list here// thus we need `next` point to next edge in adjacent list of node `u`int next;// the reverse edge (from `v` -> `u`) should be (edgeno ^ 1)// as we are adding edges sequentially// for convenience in updating residual graph Edge(){this->v = -1;this->w = -1;this->next = -1;}};// construct level graph(GL) using BFS// return true if sink is reachable from source (thus there is GL)// return false otherwisebool BFS(int GL[], int Queue[], int adjs[], Edge edges[], const int N, const int source, const int sink){// initialize all levels as negative(-1)memset(GL, -1, N * sizeof(int));GL[source] = 0;int front = 0, rear = 0;Queue[0] = source;while (front <= rear){int u = Queue[front];front++;int e = adjs[u];while (e >= 0){int v = edges[e].v;// not set before (GL in this case can be thought as visited[])if (GL[v] < 0 && edges[e].w > 0){GL[v] = GL[u] + 1;rear++;assert (rear < N);Queue[rear] = v;}// set as next node in adjacent liste = edges[e].next;}}return GL[sink] > 0;}int DFS(int u, int low, int GL[], int adjs[], Edge edges[], const int sink){// if reached sink, then return the narrowest capacity in the path (low)if (u == sink || !low){return low;}// otherwise recursively check each adjacent edge (thus reach next node)int total = 0;int e = adjs[u];while (e >= 0 && low){int v = edges[e].v;// v is at next level in level graphif (GL[v] == GL[u] + 1 && edges[e].w > 0){int flow = DFS(v, min(low, edges[e].w), GL, adjs, edges, sink);// if can reach sink (because otherwise `flow` will be 0)if (flow > 0){// update residual graphedges[e].w -= flow;edges[e ^ 1].w += flow;low -= flow;total += flow;}}e = edges[e].next;}// optimization, no need to try this node any more if it has no outgoing flowsif (total == 0){GL[u] = -1;}return total;}// for more about dinic: http://comzyh.com/blog/archives/568/int dinic(int GL[], int Queue[], int adjs[], Edge edges[], const int N, const int source, const int sink){int maxflow = 0;// construct level graph G_L using BFS// until sink cannot be reached from sourcewhile (BFS(GL, Queue, adjs, edges, N, source, sink)){// search a blocking flow (augment path) using DFS// int flow = DFS(source, numeric_limits<int>::max(), GL, adjs, edges, sink);// while (flow > 0)// {// maxflow += flow;// flow = DFS(source, numeric_limits<int>::max(), GL, adjs, edges, sink);// }maxflow += DFS(source, INF, GL, adjs, edges, sink);}return maxflow;}void addEdge(Edge edges[], int &edgeno, int adjs[], int u, int v, int w){// u -> v, capacity: wedges[edgeno].v = v;edges[edgeno].w = w;edges[edgeno].next = adjs[u];adjs[u] = edgeno;edgeno++;// v -> u, capacity: 0edges[edgeno].v = u;edges[edgeno].w = 0;edges[edgeno].next = adjs[v];adjs[v] = edgeno;edgeno++;}// see http://ycool.com/post/zhhrrm6 for how to construct networkint main(){Edge edges[100000];int edgeno = 0;int adjs[MAXN];int GL[MAXN];int Queue[MAXN];int M, N;scanf("%d%d", &M, &N);memset(adjs, -1, (N + 2) * sizeof(int));int pighouses[MAXM], customer[MAXM];for (int i = 0; i < M; ++i){scanf("%d", &pighouses[i]);customer[i] = -1;}const int source = 0, sink = N + 1;for (int i = 1; i <= N; ++i){int A, K, B;scanf("%d", &A);for (int j = 0; j < A; ++j){scanf("%d", &K);K--;// if customer i is the first customer that has key to pig house K// then draw a line from source to i with capacity of pig house Kif (customer[K] < 0){addEdge(edges, edgeno, adjs, source, i, pighouses[K]);}// else draw a line from customer[K] to customer i with unlimited capacityelse{int u = customer[K];addEdge(edges, edgeno, adjs, u, i, INF);}customer[K] = i;}scanf("%d", &B);// add edges from i to sinkaddEdge(edges, edgeno, adjs, i, sink, B);}// assert(edgeno <= MAXN + MAXN - 1 + MAXN);// printf("[debug]edges(%d):\n", edgeno);// for (int i = 0; i < edgeno; ++i)// {// printf("i: %d, v: %d, w: %d, next: %d, reverse: %d\n", i, edges[i].v, edges[i].w, edges[i].next, i ^ 1);// }N += 2;printf("%d\n", dinic(GL, Queue, adjs, edges, N, source, sink));return 0;}