HDU – 2883 kebab （最大流）
分类：ACM-图论-网络流ACM-图论
题目大意：有一个烤肉老板，，每个单位时间可以完成M的烤肉 现在有N位客人，给出每位客人来的时间，走的时间，烤肉的数量和每串烤肉所需的单位时间 问这个老板能否完成每位客人的需求
解题思路：这题和HDU 3572相似，但又不能像那题那样做，因为这题时间长度有点大 所以将时间区间当成一个点，将该区间连向超级汇点，容量为区间长度*M 将所有客人连向超级源点，容量为烤肉数量＊每串烤肉所需时间 接下来的难点就是怎么将客人和时间区间连起来了 如果时间区间在客人来的时间和走的时间这段区间内，就表明这段时间可以用来帮客人烤肉，所以可以连接，容量为INF 这样图就建好了 附上大神的详细题解 详细题解
;Edge {int from, to, cap, flow;Edge() {}Edge(int from, int to, int cap, int flow): from(from), to(to), cap(cap), flow(flow) {}};struct ISAP {int p[N], num[N], cur[N], d[N];int t, s, n, m;bool vis[N];vector<int> G[N];vector<Edge> edges;void init(int n) {this->n = n;for (int i = 0; i <= n; i++) {G[i].clear();d[i] = INF;}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;d[t] = 0;vis[t] = 1;Q.push(t);while (!Q.empty()) {int u = Q.front();Q.pop();for (int i = 0; i < G[u].size(); i++) {Edge &e = edges[G[u][i] ^ 1];if (!vis[e.from] && e.cap > e.flow) {vis[e.from] = true;d[e.from] = d[u] + 1;Q.push(e.from);}}}return vis[s];}int Augment() {int u = t, flow = INF;while (u != s) {Edge &e = edges[p[u]];flow = min(flow, e.cap – e.flow);u = edges[p[u]].from;}u = t;while (u != s) {edges[p[u]].flow += flow;edges[p[u] ^ 1].flow -= flow;u = edges[p[u]].from;}return flow;}int Maxflow(int s, int t) {this->s = s; this->t = t;int flow = 0;BFS();if (d[s] > n)return 0;memset(num, 0, sizeof(num));memset(cur, 0, sizeof(cur));for (int i = 0; i < n; i++)if (d[i] < INF)num[d[i]]++;int u = s;while (d[s] <= n) {if (u == t) {flow += Augment();u = s;}bool ok = false;for (int i = cur[u]; i < G[u].size(); i++) {Edge &e = edges[G[u][i]];if (e.cap > e.flow && d[u] == d[e.to] + 1) {ok = true;p[e.to] = G[u][i];cur[u] = i;u = e.to;break;}}if (!ok) {int Min = n;for (int i = 0; i < G[u].size(); i++) {Edge &e = edges[G[u][i]];if (e.cap > e.flow)Min = min(Min, d[e.to]);}if (–num[d[u]] == 0)break;num[d[u] = Min + 1]++;cur[u] = 0;if (u != s)u = edges[p[u]].from;}}return flow;}};ISAP isap;int S[N], E[N], num[N], T[N], All[N];int n, m;void solve() {int t, cnt = 0, s = 0, Sum = 0;for (int i = 1; i <= n; i++) {scanf(“%d%d%d%d”, &S[i], &num[i], &E[i], &T[i]);All[cnt++] = S[i];All[cnt++] = E[i];Sum += num[i] * T[i];}sort(All, All + cnt);cnt = unique(All, All + cnt) – All;t = n + cnt + 1;isap.init(t);for (int i = 1; i <= n; i++)isap.AddEdge(s, i, num[i] * T[i]);for (int i = 1; i < cnt; i++) {isap.AddEdge(i + n, t, (All[i] – All[i – 1]) * m);for (int j = 1; j <= n; j++) {if (S[j] <= All[i – 1] && E[j] >= All[i]) {isap.AddEdge(j, i + n, INF);}}}if (Sum == isap.Maxflow(s, t))printf(“Yes\n”);elseprintf(“No\n”);}int main() {while (scanf(“%d%d”, &n, &m) != EOF) {solve();}return 0;}
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