-
Notifications
You must be signed in to change notification settings - Fork 341
Commit
This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository.
- Loading branch information
Showing
6 changed files
with
598 additions
and
25 deletions.
There are no files selected for viewing
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,136 @@ | ||
| /* | ||
| * AOJ 2230: How to Create a Good Game | ||
| * 题意：给出一个带权有向无环图，在保证起点到终点的最长路不变前提下，可增加一些边的权。求最大能增加的边权总和。 | ||
| * 类型：最小费 | ||
| * 算法：输入的边两端连接容量INF花费为负边权的边，Dijkstra求最长路。从终点到起点连接容量无穷、花费为最长路的边。构造源s汇t，s与入度大于出度的点连容量度差花费的本容量1花费0的边，出度大的向t连类似的边。求得的最小费为所有圈的正权和，再加上负权和（输入边权和的相反数）即为答案。 | ||
| */ | ||
|
|
||
| #include <cstdio> | ||
| #include <cstring> | ||
| #include <algorithm> | ||
| #include <queue> | ||
|
|
||
| using namespace std; | ||
|
|
||
| typedef pair<int, int> pii; | ||
|
|
||
| const int INF = 0x3f3f3f3f; | ||
|
|
||
| int in[110], out[110]; | ||
|
|
||
| struct Edge { | ||
| Edge() {} | ||
|
|
||
| Edge(int _v, int _cap, int _cost, int _rev) : v(_v), cap(_cap), cost(_cost), rev(_rev) {} | ||
|
|
||
| int v, cap, cost, rev; | ||
| }; | ||
|
|
||
| vector<Edge> e[110]; | ||
| int n; | ||
| int h[110], d[110]; | ||
| int pv[110], pe[110]; | ||
| priority_queue<pii, vector<pii>, greater<pii> > pq; | ||
|
|
||
| void AddEdge(int u, int v, int cap, int cost) { | ||
| e[u].push_back(Edge(v, cap, cost, e[v].size())); | ||
| e[v].push_back(Edge(u, 0, -cost, e[u].size() - 1)); | ||
| } | ||
|
|
||
| int MinCostFlow(int s, int t, int f) { | ||
| int ans = 0; | ||
| memset(h, 0, sizeof(h)); | ||
| while (f > 0) { | ||
| memset(d, 0x3f, sizeof(d)); | ||
| d[s] = 0; | ||
| pq.push(make_pair(0, s)); | ||
| while (!pq.empty()) { | ||
| int u = pq.top().second; | ||
| int pre = pq.top().first; | ||
| pq.pop(); | ||
| if (d[u] < pre) continue; | ||
| for (int i = 0; i < e[u].size(); ++i) { | ||
| Edge &te = e[u][i]; | ||
| int v = te.v; | ||
| if (te.cap > 0 && d[v] > d[u] + h[u] - h[v] + te.cost) { | ||
| d[v] = d[u] + h[u] - h[v] + te.cost; | ||
| pv[v] = u; | ||
| pe[v] = i; | ||
| pq.push(make_pair(d[v], v)); | ||
| } | ||
| } | ||
| } | ||
|
|
||
| if (d[t] == INF) { | ||
| return -1; | ||
| } | ||
| for (int i = 0; i < n; ++i) { | ||
| h[i] += d[i]; | ||
| } | ||
|
|
||
| int cur = f; | ||
| for (int i = t; i != s; i = pv[i]) { | ||
| cur = min(cur, e[pv[i]][pe[i]].cap); | ||
| } | ||
| ans += cur * h[t]; | ||
| f -= cur; | ||
| for (int i = t; i != s; i = pv[i]) { | ||
| Edge &te = e[pv[i]][pe[i]]; | ||
| te.cap -= cur; | ||
| e[i][te.rev].cap += cur; | ||
| } | ||
| } | ||
| return ans; | ||
| } | ||
|
|
||
| int main() { | ||
| int m; | ||
| scanf("%d%d", &n, &m); | ||
| int s = n, t = n + 1; | ||
| int sum = 0; | ||
| for (int i = 0; i < m; ++i) { | ||
| int u, v, c; | ||
| scanf("%d%d%d", &u, &v, &c); | ||
| AddEdge(u, v, INF, -c); | ||
| sum += c; | ||
| ++in[v]; | ||
| ++out[u]; | ||
| } | ||
| int degree = 0; | ||
| for (int i = 0; i < n; ++i) { | ||
| if (in[i] > out[i]) { | ||
| degree += in[i] - out[i]; | ||
| } | ||
| } | ||
|
|
||
| memset(d, 0x3f, sizeof(d)); | ||
| d[0] = 0; | ||
| pq.push(make_pair(0, 0)); | ||
| while (!pq.empty()) { | ||
| int u = pq.top().second; | ||
| int pre = pq.top().first; | ||
| pq.pop(); | ||
| if (d[u] < pre) continue; | ||
| for (int i = 0; i < e[u].size(); ++i) { | ||
| Edge &te = e[u][i]; | ||
| int v = te.v; | ||
| if (te.cap > 0 && d[v] > d[u] + te.cost) { | ||
| d[v] = d[u] + te.cost; | ||
| pq.push(make_pair(d[v], v)); | ||
| } | ||
| } | ||
| } | ||
|
|
||
| AddEdge(n - 1, 0, INF, -d[n - 1]); | ||
| for (int i = 0; i < n; ++i) { | ||
| if (in[i] > out[i]) { | ||
| AddEdge(s, i, in[i] - out[i], 0); | ||
| } else { | ||
| AddEdge(i, t, out[i] - in[i], 0); | ||
| } | ||
| } | ||
| n += 2; | ||
| int ans = MinCostFlow(s, t, degree) - sum; | ||
| printf("%d\n", ans); | ||
| return 0; | ||
| } |
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,109 @@ | ||
| /* | ||
| * AOJ 2266: Cache Strategy | ||
| * 题意：有n个重量不同的球和m个框，再给出一个可重复的放球序列。依照序列顺序将此刻不在框内的球放入，若选择放入的框已有球，则需花费新球数量的重量来替换。求最小总花费。 | ||
| * 类型：最小费 | ||
| * 算法：将输入序列相邻相等的驱去重后，如每次都替换，则总花费为新序列的重量和。在序列中，两个相同的球之间的左闭右开序列区间如果放在同一框内，则总花费可减去该球重量。建图时可将上述区间的起点和终点建容量1花费负重量的边，再用容量为INF花费0的边将整个图连起来。求最小费用流即为可以节约的最大值的相反数，加上预处理的原始总重量即为答案。因为图有负环，所以求最短增广路不能用Dijkstra，改用Bellman-Ford。 | ||
| */ | ||
|
|
||
| #include <cstdio> | ||
| #include <cstring> | ||
| #include <algorithm> | ||
| #include <queue> | ||
|
|
||
| using namespace std; | ||
|
|
||
| typedef pair<int, int> pii; | ||
|
|
||
| const int INF = 0x3f3f3f3f; | ||
|
|
||
| int a[10010], w[10010]; | ||
| int pre[10010]; | ||
|
|
||
| struct Edge { | ||
| Edge() {} | ||
|
|
||
| Edge(int _v, int _cap, int _cost, int _rev) : v(_v), cap(_cap), cost(_cost), rev(_rev) {} | ||
|
|
||
| int v, cap, cost, rev; | ||
| }; | ||
|
|
||
| vector<Edge> e[10010]; | ||
| int n; | ||
| int d[10010]; | ||
| int pv[10010], pe[10010]; | ||
|
|
||
| void AddEdge(int u, int v, int cap, int cost) { | ||
| e[u].push_back(Edge(v, cap, cost, e[v].size())); | ||
| e[v].push_back(Edge(u, 0, -cost, e[u].size() - 1)); | ||
| } | ||
|
|
||
| int MinCostFlow(int s, int t, int f) { | ||
| int ans = 0; | ||
| while (f > 0) { | ||
| memset(d, 0x3f, sizeof(d)); | ||
| d[s] = 0; | ||
| bool update = true; | ||
| while (update) { | ||
| update = false; | ||
| for (int i = 0; i < n; ++i) { | ||
| if (d[i] == INF) continue; | ||
| for (int j = 0; j < e[i].size(); ++j) { | ||
| Edge &te = e[i][j]; | ||
| if (te.cap > 0 && d[te.v] > d[i] + te.cost) { | ||
| d[te.v] = d[i] + te.cost; | ||
| pv[te.v] = i; | ||
| pe[te.v] = j; | ||
| update = true; | ||
| } | ||
| } | ||
| } | ||
| } | ||
|
|
||
| if (d[t] == INF) { | ||
| return -1; | ||
| } | ||
|
|
||
| int cur = f; | ||
| for (int i = t; i != s; i = pv[i]) { | ||
| cur = min(cur, e[pv[i]][pe[i]].cap); | ||
| } | ||
| ans += cur * d[t]; | ||
| f -= cur; | ||
| for (int i = t; i != s; i = pv[i]) { | ||
| Edge &te = e[pv[i]][pe[i]]; | ||
| te.cap -= cur; | ||
| e[i][te.rev].cap += cur; | ||
| } | ||
| } | ||
| return ans; | ||
| } | ||
|
|
||
| int main() { | ||
| int M, N, K, ans = 0; | ||
| scanf("%d%d%d", &M, &N, &K); | ||
| for (int i = 0; i < N; ++i) { | ||
| scanf("%d", &w[i]); | ||
| } | ||
| for (int i = 0; i < K; ++i) { | ||
| scanf("%d", &a[i]); | ||
| --a[i]; | ||
| } | ||
| K = unique(a, a + K) - a; | ||
|
|
||
| memset(pre, -1, sizeof(pre)); | ||
| for (int i = 0; i < K; ++i) { | ||
| ans += w[a[i]]; | ||
| if (pre[a[i]] != -1) { | ||
| AddEdge(pre[a[i]], i - 1, 1, -w[a[i]]); | ||
| } | ||
| pre[a[i]] = i; | ||
| } | ||
| for (int i = 1; i < K; ++i) { | ||
| AddEdge(i - 1, i, M - 1, 0); | ||
| } | ||
| n = K; | ||
| ans += MinCostFlow(0, n - 1, M - 1); | ||
| printf("%d\n", ans); | ||
|
|
||
| return 0; | ||
| } |
Oops, something went wrong.