Ombrophobic Bovines
FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They have decided to put a rain siren on the farm to let them know when rain is approaching. They intend to create a rain evacuation plan so that all the cows can get to shelter before the rain begins. Weather forecasting is not always correct, though. In order to minimize false alarms, they want to sound the siren as late as possible while still giving enough time for all the cows to get to some shelter.
The farm has F (1 <= F <= 200) fields on which the cows graze. A set of P (1 <= P <= 1500) paths connects them. The paths are wide, so that any number of cows can traverse a path in either direction.
Some of the farm's fields have rain shelters under which the cows can shield themselves. These shelters are of limited size, so a single shelter might not be able to hold all the cows. Fields are small compared to the paths and require no time for cows to traverse.
Compute the minimum amount of time before rain starts that the siren must be sounded so that every cow can get to some shelter.
Input
* Line 1: Two space-separated integers: F and P
* Lines 2..F+1: Two space-separated integers that describe a field. The first integer (range: 0..1000) is the number of cows in that field. The second integer (range: 0..1000) is the number of cows the shelter in that field can hold. Line i+1 describes field i.
* Lines F+2..F+P+1: Three space-separated integers that describe a path. The first and second integers (both range 1..F) tell the fields connected by the path. The third integer (range: 1..1,000,000,000) is how long any cow takes to traverse it.
Output
* Line 1: The minimum amount of time required for all cows to get under a shelter, presuming they plan their routes optimally. If it not possible for the all the cows to get under a shelter, output "-1".
Sample Input
3 4
7 2
0 4
2 6
1 2 40
3 2 70
2 3 90
1 3 120
Sample Output
110
Hint
OUTPUT DETAILS:
In 110 time units, two cows from field 1 can get under the shelter in that field, four cows from field 1 can get under the shelter in field 2, and one cow can get to field 3 and join the cows from that field under the shelter in field 3. Although there are other plans that will get all the cows under a shelter, none will do it in fewer than 110 time units.
Source
题目类型：最大流+二分枚举+Floyd
算法分析：由于所有点都在一个点集中，可能出现两个点间接相连的情况，所以先使用拆点的方法将每个点A拆成A’和A”，连一条A’至A”容量为INF的有向边。接着二分枚举回到避雨点所需要的时间mid。对于当前的mid，若任意两点A，B之间的距离小于等于mid，则从A’向B”连一条容量为INF的有向边，最后从源点向每个i’点连一条容量为i点当前牛数的有向边，每个点i”向汇点连一条容量为i点可容纳牛数的有向边。最后跑一个最大流并依据结果调整二分枚举的方向即可。注意最短路求解时需要使用long long！！！
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/************************************************* Author :supermaker Created Time :2016/3/27 8:56:34 File Location :C:\Users\abcd\Desktop\TheEternalPoet **************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 400 + 66; struct Node { int to, nxt, c, f; }; Node edge[maxn*maxn+3*maxn]; int head[maxn], edgelen; int dep[maxn], cnt[maxn], cur[maxn]; void Init () { memset (edge, -1, sizeof (edge)); memset (head, -1, sizeof (head)); edgelen = 0; } void AddEdge (int u, int v, int w, int rw = 0) { edge[edgelen].to = v, edge[edgelen].c = w, edge[edgelen].f = 0; edge[edgelen].nxt = head[u], head[u] = edgelen++; edge[edgelen].to = u, edge[edgelen].c = rw, edge[edgelen].f = 0; edge[edgelen].nxt = head[v], head[v] = edgelen++; } void bfs (int s, int e) { memset (dep, -1, sizeof (dep)); memset (cnt, 0, sizeof (cnt)); dep[e] = 0; cnt[dep[e]]++; queue <int> qu; qu.push (e); while (!qu.empty ()) { int u = qu.front (); qu.pop (); for (int i = head[u]; i != -1; i = edge[i].nxt) { int v = edge[i].to; if (dep[v] != -1) continue; qu.push (v); dep[v] = dep[u] + 1; cnt[dep[v]]++; } } } int sta[maxn]; int MaxFlow (int s, int e, int n) { bfs (s, e); int res = 0, top = 0, u = s; memcpy (cur, head, sizeof (head)); while (dep[s] < n) { if (u == e) { int minval = INF, minval_pos = -1; for (int i = 0; i < top; i++) if (minval > edge[sta[i]].c - edge[sta[i]].f) { minval = edge[sta[i]].c - edge[sta[i]].f; minval_pos = i; } for (int i = 0; i < top; i++) { edge[sta[i]].f += minval; edge[sta[i]^1].f -= minval; } res += minval; top = minval_pos; u = edge[sta[top]^1].to; } else { bool is_have = false; for (int i = cur[u]; i != -1; i = edge[i].nxt) { int v = edge[i].to; if (edge[i].c - edge[i].f && dep[v] + 1 == dep[u]) { cur[u] = i; sta[top++] = cur[u]; u = v; is_have = true; break; } } if (is_have) continue; int minval = n; for (int i = head[u]; i != -1; i = edge[i].nxt) if (edge[i].c - edge[i].f && dep[edge[i].to] < minval) { minval = dep[edge[i].to]; cur[u] = i; } cnt[dep[u]]--; if (!cnt[dep[u]]) return res; dep[u] = minval + 1; cnt[dep[u]]++; if (u != s) u = edge[sta[--top]^1].to; } } return res; } LL g[maxn][maxn], dis[maxn][maxn]; int F, P, tot, now[maxn], cap[maxn]; void floyd () { for (int i = 1; i <= F; i++) for (int j = 1; j <= F; j++) dis[i][j] = g[i][j]; for (int k = 1; k <= F; k++) for (int i = 1; i <= F; i++) for (int j = 1; j <= F; j++) { if (i == k || j == k) continue; dis[i][j] = min (dis[i][j], dis[i][k] + dis[k][j]); } } void SS (LL val) { Init (); for (int i = 1; i <= F; i++) AddEdge (i, F + i, INF); for (int i = 1; i <= F; i++) for (int j = 1; j <= F; j++) if (dis[i][j] <= val) AddEdge (i, F + j, INF); for (int i = 1; i <= F; i++) AddEdge (0, i, now[i]); for (int i = 1; i <= F; i++) AddEdge (F + i, 2 * F + 1, cap[i]); } int main() { //CFF; //CPPFF; while (scanf ("%d%d", &F, &P) != EOF) { tot = 0; for (int i = 1; i <= F; i++) { scanf ("%d%d", &now[i], &cap[i]); tot += now[i]; } for (int i = 0; i < maxn; i++) for (int j = 0; j < maxn; j++) { if (i == j) g[i][j] = 0; else g[i][j] = (LL) 2e12; } for (int i = 1; i <= P; i++) { LL u, v, w; scanf ("%lld%lld%lld", &u, &v, &w); if (g[u][v] > w) g[u][v] = g[v][u] = w; } floyd (); LL low = 0, high = 0, mid; for (int i = 1; i <= F; i++) for (int j = 1; j <= F; j++) if (dis[i][j] < (LL) 2e12) high = max (high, dis[i][j]); LL ans = -1; while (low <= high) { mid = (low + high) >> 1; //DB (mid); SS (mid); int res = MaxFlow (0, F * 2 + 1, F * 2 + 2); if (res >= tot) { ans = mid; high = mid - 1; } else low = mid + 1; } printf ("%lld\n", ans); } return 0; } |
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