首页 > 代码库 > POJ 2391.Ombrophobic Bovines 解题报告
POJ 2391.Ombrophobic Bovines 解题报告
实际上是求最短的避雨时间。
首先将每个点拆成两个,一个连接源点,一个连接汇点,连接源点的点的容量为当前单的奶牛数,连接汇点的点为能容纳的奶牛数。
floyd求任意两点互相到达的最短时间,二分最长时间,最大流判断是否可行。
注意路径时间会超过int
/* 最大流SAP 邻接表 思路:基本源于FF方法,给每个顶点设定层次标号,和允许弧。 优化: 1、当前弧优化(重要)。 1、每找到以条增广路回退到断点(常数优化)。 2、层次出现断层,无法得到新流(重要)。 时间复杂度(m*n^2)*/#include <iostream>#include <cstdio>#include <cstring>#define ms(a,b) memset(a,b,sizeof a)using namespace std;const int INF = 500;long long G[INF][INF];struct node { int v, c, next;} edge[INF*INF * 4];long long pHead[INF*INF], SS, ST, nCnt;void addEdge (int u, int v, int c) { edge[++nCnt].v = v; edge[nCnt].c = c, edge[nCnt].next = pHead[u]; pHead[u] = nCnt; edge[++nCnt].v = u; edge[nCnt].c = 0, edge[nCnt].next = pHead[v]; pHead[v] = nCnt;}int SAP (int pStart, int pEnd, int N) { int numh[INF], h[INF], curEdge[INF], pre[INF]; int cur_flow, flow_ans = 0, u, neck, i, tmp; ms (h, 0); ms (numh, 0); ms (pre, -1); for (i = 0; i <= N; i++) curEdge[i] = pHead[i]; numh[0] = N; u = pStart; while (h[pStart] <= N) { if (u == pEnd) { cur_flow = 1e9; for (i = pStart; i != pEnd; i = edge[curEdge[i]].v) if (cur_flow > edge[curEdge[i]].c) neck = i, cur_flow = edge[curEdge[i]].c; for (i = pStart; i != pEnd; i = edge[curEdge[i]].v) { tmp = curEdge[i]; edge[tmp].c -= cur_flow, edge[tmp ^ 1].c += cur_flow; } flow_ans += cur_flow; u = neck; } for ( i = curEdge[u]; i != 0; i = edge[i].next) if (edge[i].c && h[u] == h[edge[i].v] + 1) break; if (i != 0) { curEdge[u] = i, pre[edge[i].v] = u; u = edge[i].v; } else { if (0 == --numh[h[u]]) continue; curEdge[u] = pHead[u]; for (tmp = N, i = pHead[u]; i != 0; i = edge[i].next) if (edge[i].c) tmp = min (tmp, h[edge[i].v]); h[u] = tmp + 1; ++numh[h[u]]; if (u != pStart) u = pre[u]; } } return flow_ans;}long long m, n, x, y,sum,c;int in[INF], out[INF];bool check (long long tem) { nCnt = 1; SS = 2 * n + 1, ST = 2 * n + 2; memset (pHead, 0, sizeof pHead); for (int i = 1; i <= n; i++) { if(out[i]) addEdge (SS, i, out[i]); for (int j =1; j <= n; j++) if (G[i][j] <= tem&&G[i][j]!=-1) addEdge (i, j+n, 5000); } for (int i = 1; i <= n; i++) if(in[i])addEdge (i + n, ST, in[i]); int ans = SAP (SS, ST, ST); if (ans == sum) return 1; return 0;}int main() { /* 建图,前向星存边,表头在pHead[],边计数 nCnt. SS,ST分别为源点和汇点 */ ms (G, -1); cin>>n>>m; for (int i = 1; i <= n; i++) { cin>>out[i]>>in[i]; sum += out[i]; } long long l = 0x7fffffffffffffff, r = 0; for (int i = 1; i <= n; i++) G[i][i] = 0; for (int i = 1; i <= m; i++) { cin>>x>>y>>c; if(G[x][y]>0) G[x][y] = G[y][x] = min(c,G[x][y]); else G[x][y]=G[y][x]=c; l = min (l, c), r = max (r, c); } for (int t = 1; t <= n; t++) for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) { if (G[i][t] == -1 || G[t][j] == -1) continue; if (G[i][j] == -1 || G[i][j] > G[i][t] + G[t][j]) G[i][j] = G[i][t] + G[t][j], l = min (l, G[i][j]), r = max (r, G[i][j]); } long long last = -1,mid; while (l <= r) { mid = (l + r) >> 1; if (check (mid) ) { last = mid; r = mid - 1; } else l = mid + 1; } cout<<last; return 0;}
POJ 2391.Ombrophobic Bovines 解题报告
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。