After the end of the truck drivers‘ strike, you and the rest of Nlog?nia logistics specialists now have the task of planning the refueling of the gas stations in the city. For this, we collected information on stocks of R refineries and about the demands of P
gas stations. In addition, there are contractual restrictions that some refineries cannot supply some gas stations; When a refinery can provide a station, the shorter route to transport fuel from one place to another is known.
The experts‘ task is to minimize the time all stations are supplied, satisfying their demands. The refineries have a sufficiently large amount of trucks, so that you can assume that each truck will need to make only one trip from a refinery to a gas station. The capacity of each truck is greater than the demand of any gas station, but it may be necessary to use more than one refinery.
Input
The first line of the input contains three integers P,R,C
, respectively the number of gas stations, the number of refineries and the number of pairs of refineries and gas stations whose time will be given (1≤P,R≤1000; 1≤C≤20000). The second line contains P integers Di (1≤Di≤104), representing the demands in liters of gasoline of the gas stations i=1,2,…,P, in that order. The third line contains R integers Ei (1≤Ei≤104), representing stocks, in liters of gasoline, of refineries i=1,2,…,R, in that order. Finally, the latest C lines describe course times, in minutes, between stations and refineries. Each of these rows contains three integers, I,J,T (1≤I≤P; 1≤J≤R; 1≤T≤106), where I is the ID of a post, J is the ID of a refinery and T is the time in the course of a refinery truck J to I. No pair (J,I)
repeats. Not all pairs are informed; If a pair is not informed, contractual restrictions prevents the refinery from supplying the station.
Output
Print an integer that indicates the minimum time in minutes for all stations to be completely filled up. If this is not possible, print ?1.
Examples
Input
3 2 520 10 1030 201 1 22 1 12 2 33 1 43 2 5
Output
4
Input
3 2 520 10 1025 301 1 32 1 12 2 43 1 23 2 5
Output
5
Input
4 3 910 10 10 2010 15 301 1 11 2 12 1 32 2 23 1 103 2 104 1 14 2 24 3 30
Output
-1
Input
1 2 24030 101 1 1001 2 200
Output
200
有点像费用流，但是复杂度过不去；
由于我们要求最短时间，考虑二分答案；
对于此时的时间，显然只有<=x的边才能相连,
并且该边的流量设为inf；
然后设立源点，汇点跑一下最大流，看是否满流即可；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 300005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == ‘-‘) f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
int n, m;
int st, ed;
struct node {
int u, v, nxt, w;
}edge[maxn << 1];
int head[maxn], cnt;
void addedge(int u, int v, int w) {
edge[cnt].u = u; edge[cnt].v = v; edge[cnt].nxt = head[u];
edge[cnt].w = w; head[u] = cnt++;
}
int rk[maxn];
int bfs() {
queue<int>q;
ms(rk);
rk[st] = 1;
q.push(st);
while (!q.empty()) {
int tmp = q.front(); q.pop();
for (int i = head[tmp]; i != -1; i = edge[i].nxt) {
int to = edge[i].v;
if (rk[to] || edge[i].w <= 0)continue;
rk[to] = rk[tmp] + 1; q.push(to);
}
}
return rk[ed];
}
int dfs(int u, int flow) {
if (u == ed)return flow;
int add = 0;
for (int i = head[u]; i != -1 && add < flow; i = edge[i].nxt) {
int v = edge[i].v;
if (rk[v] != rk[u] + 1 || !edge[i].w)continue;
int tmpadd = dfs(v, min(edge[i].w, flow - add));
if (!tmpadd) { rk[v] = -1; continue; }
edge[i].w -= tmpadd; edge[i ^ 1].w += tmpadd;
add += tmpadd;
}
return add;
}
int ans;
void dinic() {
while (bfs())ans += dfs(st, inf);
}
int P, R, C;
int D[maxn], E[maxn], T;
int sum;
struct nd {
int u, v, w;
}e[maxn];
bool chk(int x) {
ms(edge); memset(head, -1, sizeof(head)); cnt = 0;
ms(rk);
ans = 0;
st = 0; ed = P + R + 2;
for (int i = 1; i <= R; i++)addedge(st, i, E[i]), addedge(i, st, 0);
for (int i = 1; i <= P; i++)addedge(i + R, ed, D[i]), addedge(ed, i + R, 0);
for (int i = 1; i <= C; i++) {
if (x >= e[i].w) {
addedge(e[i].v, e[i].u + R, inf); addedge(e[i].u + R, e[i].v, 0);
}
}
dinic();
if (ans == sum)return true;
return false;
}
int main()
{
// ios::sync_with_stdio(0);
memset(head, -1, sizeof(head));
P = rd(); R = rd(); C = rd();
for (int i = 1; i <= P; i++) {
D[i] = rd(); sum += D[i];// gas stations
}
for (int i = 1; i <= R; i++)E[i] = rd();
for (int i = 1; i <= C; i++)e[i].u = rd(), e[i].v = rd(), e[i].w = rd();
bool fg = 0;
int l = 0, r = 1e7 + 1;
int as = 0;
while (l <= r) {
int mid = (l + r) / 2;
if (chk(mid)) {
r = mid - 1; as = mid; fg = 1;
}
else l = mid + 1;
}
if (!fg)cout << -1 << endl;
else cout << as << endl;
return 0;
}
原文地址：https://www.cnblogs.com/zxyqzy/p/10353227.html