【BZOJ1738】[Usaco2005 mar]Ombrophobic Bovines 发抖的牛【二分】【Floyd】【最大流】
2016-04-04 21:41
417 查看
【题目链接】
同【POJ2391题解】
同【POJ2391题解】
/* Pigonometry */ #include <cstdio> #include <algorithm> using namespace std; typedef long long LL; const int maxn = 1005, maxm = 100005, maxq = 10000, inf = 0x3f3f3f3f; const LL infinf = (LL)inf * inf; int n, m, cur[maxn], head[maxn], cnt, depth[maxn], bg, ed, q[maxq], maxflow, A[maxn], B[maxn]; LL map[maxn][maxn]; struct _edge { int v, w, next; } g[maxm]; inline int iread() { int f = 1, x = 0; char ch = getchar(); for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1; for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0'; return f * x; } inline void add(int u, int v, int w) { g[cnt] = (_edge){v, w, head[u]}; head[u] = cnt++; } inline void insert(int u, int v, int w) { add(u, v, w); add(v, u, 0); } inline bool bfs() { for(int i = 0; i <= ed; i++) depth[i] = -1; int h = 0, t = 0, u, i; depth[q[t++] = bg] = 1; while(h != t) for(i = head[u = q[h++]]; ~i; i = g[i].next) if(g[i].w && !~depth[g[i].v]) { depth[g[i].v] = depth[u] + 1; if(g[i].v == ed) return 1; q[t++] = g[i].v; } return 0; } inline int dfs(int x, int flow) { if(x == ed) return flow; int left = flow; for(int i = cur[x]; ~i; i = g[i].next) if(g[i].w && depth[g[i].v] == depth[x] + 1) { int tmp = dfs(g[i].v, min(left, g[i].w)); left -= tmp; g[i].w -= tmp; g[i ^ 1].w += tmp; if(g[i].w) cur[x] = i; if(!left) return flow; } if(left == flow) depth[x] = -1; return flow - left; } inline bool check(LL tim) { for(int i = 0; i <= ed; i++) head[i] = -1; cnt = 0; for(int i = 1; i <= n; i++) { insert(bg, i, A[i]); insert(i + n, ed, B[i]); } for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) if(map[i][j] <= tim) insert(i, n + j, inf); int ans = maxflow; while(bfs()) { for(int i = 0; i <= ed; i++) cur[i] = head[i]; ans -= dfs(bg, inf); } return ans; } int main() { n = iread(); m = iread(); bg = 0; ed = (n << 1) + 1; for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) if(i ^ j) map[i][j] = infinf; int maxb = 0; for(int i = 1; i <= n; i++) { A[i] = iread(); B[i] = iread(); maxflow += A[i]; maxb += B[i]; } if(maxflow > maxb) { printf("-1\n"); return 0; } LL maxw = 0; for(int i = 1; i <= m; i++) { int u = iread(), v = iread(); LL w; scanf("%lld", &w); map[u][v] = min(map[u][v], w); map[v][u] = min(map[v][u], w); maxw += w; } for(int k = 1; k <= n; k++) for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) map[i][j] = min(map[i][j], map[i][k] + map[k][j]); LL l = 0, r = maxw; while(l <= r) { LL mid = l + r >> 1; if(check(mid)) l = mid + 1; else r = mid - 1; } if(r == maxw) l = -1; printf("%lld\n", l); return 0; }
相关文章推荐
- 快速排序里的学问:从猜数字开始
- 封装好的Folyd建图,C++源码
- Hdu2066(一个人的旅行)
- HDU 4898 The Revenge of the Princess’ Knight ( 2014 Multi-University Training Contest 4 )
- Search Insert Position,Search for a Range,Pow(x, n),Sqrt(x)
- Find Minimum in Rotated Sorted Array II
- 【网络流-最大流-Dinic-建模】POJ3281 Dining:Pascal 解法
- Ford_Fulkerson算法
- 重标记与前移算法
- 网络流之增广路算法
- Poj2638 网络流+最短路+二分答案
- Aizu1311 分层图最短路 (...大概)
- [LeetCode] Sqrt(x)
- [LeetCode] Pow(x, n)
- [LeetCode] Search Insert Position
- [LeetCode] Search for a Range
- [LeetCode] Search in Rotated Sorted Array
- [笔记] 网络流-最大流 POJ-1273\HDU-4240
- 最短路径 -- spfa
- PAT 1057 Stack (30)