POJ 1698 Alice's Chance（最大流+拆点）

POJ 1698 Alice's Chance

题目链接

题意：拍n部电影，每部电影要在前w星期完成，并且一周只有一些天是可以拍的，每部电影有个需要的总时间，问是否能拍完电影

思路：源点向每部电影连边，容量为d，然后每部电影对应能拍的那天连边，由于每天容量限制是1，所以进行拆点，然后连向汇点即可

代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

const int MAXNODE = 1005;
const int MAXEDGE = 100005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
int u, v;
Type cap, flow;
Edge() {}
Edge(int u, int v, Type cap, Type flow) {
this->u = u;
this->v = v;
this->cap = cap;
this->flow = flow;
}
};

struct Dinic {
int n, m, s, t;
Edge edges[MAXEDGE];
int first[MAXNODE];
int next[MAXEDGE];
bool vis[MAXNODE];
Type d[MAXNODE];
int cur[MAXNODE];
vector<int> cut;

void init(int n) {
this->n = n;
memset(first, -1, sizeof(first));
m = 0;
}
void add_Edge(int u, int v, Type cap) {
edges[m] = Edge(u, v, cap, 0);
next[m] = first[u];
first[u] = m++;
edges[m] = Edge(v, u, 0, 0);
next[m] = first[v];
first[v] = m++;
}

bool bfs() {
memset(vis, false, sizeof(vis));
queue<int> Q;
Q.push(s);
d[s] = 0;
vis[s] = true;
while (!Q.empty()) {
int u = Q.front(); Q.pop();
for (int i = first[u]; i != -1; i = next[i]) {
Edge& e = edges[i];
if (!vis[e.v] && e.cap > e.flow) {
vis[e.v] = true;
d[e.v] = d[u] + 1;
Q.push(e.v);
}
}
}
return vis[t];
}

Type dfs(int u, Type a) {
if (u == t || a == 0) return a;
Type flow = 0, f;
for (int &i = cur[u]; i != -1; i = next[i]) {
Edge& e = edges[i];
if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) {
e.flow += f;
edges[i^1].flow -= f;
flow += f;
a -= f;
if (a == 0) break;
}
}
return flow;
}

Type Maxflow(int s, int t) {
this->s = s; this->t = t;
Type flow = 0;
while (bfs()) {
for (int i = 0; i < n; i++)
cur[i] = first[i];
flow += dfs(s, INF);
}
return flow;
}

void MinCut() {
cut.clear();
for (int i = 0; i < m; i += 2) {
if (vis[edges[i].u] && !vis[edges[i].v])
cut.push_back(i);
}
}
} gao;

int t, n, day[10], d, w, vis[400];

int main() {
scanf("%d", &t);
while (t--) {
memset(vis, 0, sizeof(vis));
gao.init(1000);
scanf("%d", &n);
int sum = 0;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= 7; j++)
scanf("%d", &day[j]);
scanf("%d%d", &d, &w);
sum += d;
gao.add_Edge(0, i, d);
for (int j = 0; j < w; j++) {
for (int k = 1; k <= 7; k++) {
if (day[k]) {
gao.add_Edge(i, 20 + j * 7 + k, INF);
vis[20 + j * 7 + k] = 1;
}
}
}
}
for (int i = 21; i <= 370; i++) {
if (!vis[i]) continue;
gao.add_Edge(i, i + 350, 1);
gao.add_Edge(i + 350, 1000, INF);
}
printf("%s\n", gao.Maxflow(0, 1000) == sum ? "Yes" : "No");
}
return 0;
}