HDU-3572 Task Schedule (最大流Dinic做法)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3572
我的博客:https://acmerszq.cn
题意:
M台机器处理N个任务。对于第i个任务,工厂必须在第天或之后开始处理,处理天,并在之前或这天完成任务。一台机器只能处理一个任务,一个任务一次只能被一台机器处理,但是能在不同日期和不同机器上中断和处理。判断是否存在时间表完成所有任务。
思路:
基于天数建图。如下图:
注意:容量为m的路不要重复建路
AC代码:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2010; //点数的最大值
const int MAXM = 1200010; //边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge {
int to, next, cap, flow;
} edge[MAXM]; //注意是 MAXM
int tol;
int head[MAXN];
void init() {
tol = 2;
memset(head, -1, sizeof(head));
}
void addedge(int u, int v, int w, int rw = 0) {
edge[tol].to = v;
edge[tol].cap = w;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = rw;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}
int Q[MAXN];
int dep[MAXN], cur[MAXN], sta[MAXN];
bool bfs(int s, int t, int n) {
int front = 0, tail = 0;
memset(dep, -1, sizeof(dep[0]) * (n + 1));
dep[s] = 0;
Q[tail++] = s;
while (front < tail) {
int u = Q[front++];
for (int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
if (edge[i].cap > edge[i].flow && dep[v] == -1) {
dep[v] = dep[u] + 1;
if (v == t) return true;
Q[tail++] = v;
}
}
}
return false;
}
int dinic(int s, int t, int n) {
int maxflow = 0;
while (bfs(s, t, n)) {
for (int i = 0; i < n; i++) cur[i] = head[i];
int u = s, tail = 0;
while (cur[s] != -1) {
if (u == t) {
int tp = INF;
for (int i = tail - 1; i >= 0; i--) tp = min(tp, edge[sta[i]].cap - edge[sta[i]].flow);
maxflow += tp;
for (int i = tail - 1; i >= 0; i--) {
edge[sta[i]].flow += tp;
edge[sta[i] ^ 1].flow -= tp;
if (edge[sta[i]].cap - edge[sta[i]].flow == 0) tail = i;
}
u = edge[sta[tail] ^ 1].to;
} else if (cur[u] != -1 && edge[cur[u]].cap > edge[cur[u]].flow && dep[u] + 1 == dep[edge[cur[u]].to]) {
sta[tail++] = cur[u];
u = edge[cur[u]].to;
} else {
while (u != s && cur[u] == -1) u = edge[sta[--tail] ^ 1].to;
cur[u] = edge[cur[u]].next;
}
}
}
return maxflow;
}
bool vis[MAXN];
int main() {
// freopen("RAW/in", "r", stdin);
// freopen("RAW/out", "w", stdout);
int T;
scanf("%d", &T);
for (int _ = 1; _ <= T; _++) {
init();
memset(vis, false, sizeof(vis));
printf("Case %d: ", _);
int n, m;
scanf("%d%d", &n, &m);
int sum = 0;
for (int i = 0; i < n; i++) {
int p, s, e;
scanf("%d%d%d", &p, &s, &e);
sum += p;
addedge(0, 501 + i, p);
for (int j = s; j <= e; j++) {
addedge(501 + i, j, 1);
if (!vis[j]) {
addedge(j, 501 + n, m);
}
vis[j] = true;
}
}
int ans = dinic(0, 501 + n, n + 2 + 500);
if (ans == sum)
printf("Yes\n");
else
printf("No\n");
printf("\n");
}
return 0;
}