【BZOJ1221】[HNOI2001] 软件开发【最小费用最大流】

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【题目链接】

听说是网络流24题。。

【rausen的题解】

/* Telekinetic Forest Guard */#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 2005, maxm = 10005, maxq = 10000, inf = 0x3f3f3f3f;int n, A, B, F, FA, FB, head[maxn], cnt, bg, ed, depth[maxn], q[maxq], way[maxn];bool vis[maxn];struct _edge {int v, w, c, next;} g[maxm << 1];inline int iread() {int f = 1, x = 0; char ch = getchar();for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';return f * x;}inline void add(int u, int v, int w, int c) {g[cnt] = (_edge){v, w, c, head[u]};head[u] = cnt++;}inline void insert(int u, int v, int w, int c) {add(u, v, w, c); add(v, u, 0, -c);}inline bool spfa() {for(int i = 0; i <= ed; i++) depth[i] = inf;int h = 0, t = 0, u, i; depth[q[t++] = bg] = 0;while(h != t) for(i = head[u = q[h++]], vis[u] = 0; ~i; i = g[i].next) if(g[i].w && depth[g[i].v] > depth[u] + g[i].c) {depth[g[i].v] = depth[u] + g[i].c;way[g[i].v] = i;if(!vis[g[i].v]) vis[q[t++] = g[i].v] = 1;}return depth[ed] != inf;}inline int back() {int res = 0, flow = inf;for(int u = ed; u != bg; u = g[way[u] ^ 1].v) flow = min(flow, g[way[u]].w);for(int u = ed; u != bg; u = g[way[u] ^ 1].v) g[way[u]].w -= flow, g[way[u] ^ 1].w += flow, res += g[way[u]].c * flow;return res;}int main() {n = iread(); A = iread(); B = iread(); F = iread(); FA = iread(); FB = iread();bg = 0; ed = n + n + 1;for(int i = 0; i <= ed; i++) head[i] = -1; cnt = 0;for(int i = 1; i <= n; i++) {int x = iread();insert(bg, i, x, 0);insert(n + i, ed, x, 0);insert(bg, n + i, inf, F);if(i + A + 1 <= n) insert(i, n + i + A + 1, inf, FA);if(i + B + 1 <= n) insert(i, n + i + B + 1, inf, FB);if(i != n) insert(i, i + 1, inf, 0);}int ans = 0;while(spfa()) ans += back();printf("%d\n", ans);return 0;}


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