传送门:QAQ
题意:给你一些接受点和提供点,然后给你一些接受点到提供点的边和所需时间,求最短时间(同时运输)将所有接受点装满,输出所需时间。
思路:二分时间,然后把满足时间的点加入边,跑网络流即可。
附上代码:
#include<stdio.h>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<queue>
using namespace std;
const int maxn = 21000;
#define inf 0x3f3f3f3f
int ax[1100];
int bn[1100];
struct inst {
int x, y, w1;
};
inst ww[21000];
int n, m;
int n1, m1, k;
int sum1;
int sum2;
struct Edge {
Edge() {}
Edge(int from, int to, int cap, int flow) :from(from), to(to), cap(cap), flow(flow) {}
int from, to, cap, flow;
};
struct Dinic {
int n, m, s, t;
vector<Edge>edges;
vector<int>G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn];
void init(int n) {
this->n = n;
for (int i = 0; i <= n; i++) G[i].clear();
edges.clear();
}
void AddEdge(int from, int to, int cap) {
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, 0, 0));
m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
bool BFS() {
memset(vis, 0, sizeof(vis));
queue<int>Q;
Q.push(s);
d[s] = 0;
vis[s] = true;
while (!Q.empty()) {
int x = Q.front();
Q.pop();
for (int i = 0; i < G[x].size(); i++) {
Edge& e = edges[G[x][i]];
if (!vis[e.to] && e.cap > e.flow) {
vis[e.to] = true;
d[e.to] = d[x] + 1;
Q.push(e.to);
}
}
}
return vis[t];
}
int DFS(int x, int a) {
if (x == t || a == 0) return a;
int flow=0,f;
for (int& i = cur[x]; i < G[x].size(); i++) {
Edge& e = edges[G[x][i]];
if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
e.flow += f;
edges[G[x][i] ^ 1].flow -= f;
flow += f;
a -= f;
if (a == 0) break;
}
}
return flow;
}
int Maxflow(int s, int t) {
this->s = s, this->t = t;
int flow = 0;
while (BFS()) {
memset(cur, 0, sizeof(cur));
flow += DFS(s, inf);
}
return flow;
}
}DC;
int check(int mid) {
DC.init(2100);
int S = 0;
int T = n1 + m1 + 1;
for (int i = 1; i <= m1; i++)
{
DC.AddEdge(S, i, bn[i]);
}
for (int i = 1; i <= n1; i++) {
DC.AddEdge(m1 + i, T, ax[i]);
}
for (int i = 0; i < k; i++) {
if (ww[i].w1<=mid) {
DC.AddEdge(ww[i].y, m1 + ww[i].x, ax[ww[i].x]);
}
}
if (DC.Maxflow(S, T) >= sum1) {
return 1;
}
else return 0;
}
int main(void) {
scanf("%d%d%d", &n1, &m1, &k);
sum1 = 0;
sum2 = 0;
for (int i = 1; i <= n1; i++) {
scanf("%d", &ax[i]);
sum1 += ax[i];
}
for (int i = 1; i <= m1; i++) {
scanf("%d", &bn[i]);
sum2 += bn[i];
}
if (sum1 > sum2) {
printf("-1\n");
return 0;
}
for (int i = 0; i < k; i++) {
scanf("%d%d%d", &ww[i].x, &ww[i].y, &ww[i].w1);
}
int left = 0;
int right = 1000000;
int ans = -1;
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) {
ans = mid;
right = mid - 1;
}
else left = mid + 1;
}
printf("%d\n", ans);
return 0;
}