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题意： 有n（n<=200） 个恰好需要一天完成的任务，要求用最少的时间完成所有任务。任务可以并行完成，但必须满足一些约束，约束分为有向约束和无向约束两种，其中A->B表示A必须在B之前完成，A-B表示A和B不能在同一天晚上。输入保证约束图是将一颗树的一些边定向之后得到的。
分析： 参考紫书P297-298，写的很详细。算是一道比较复杂的树形dp了。
LRJ代码：
#include<bits/stdc++.h>
using namespace std;const int maxn = 200 + 5;
const int INF = 1000000000;struct Edge {int u, v, d; // d=1 means u->v, d=2 means v->u, d=0 means u-vEdge(int u=0, int v=0, int d=0):u(u),v(v),d(d){}
};vector<Edge> edges[maxn];
int n, root, maxlen, f[maxn], g[maxn], have_father[maxn];// maximal length of a DIRECTED path starting from u
int dfs(int u) {int ans = 0;for(int i = 0; i < edges[u].size(); i++) {int v = edges[u][i].v;if(edges[u][i].d == 1)ans = max(ans, dfs(v)+1);}return ans;
}bool read_data() {bool have_data = false;int a, b;n = 0;for(int i = 0; i < maxn; i++) edges[i].clear();memset(have_father, 0, sizeof(have_father));while(cin >> a && a){string str;have_data = true;if(a > n) n = a;while(cin >> str && str != "0"){int len = str.length();char dir = str[len-1];if(dir == 'd' || dir == 'u') str = str.substr(0, len-1);stringstream ss(str);ss >> b; // b is a's sonif(b > n) n = b;have_father[b] = 1;if(dir == 'd'){edges[a].push_back(Edge(a, b, 1)); // forwardedges[b].push_back(Edge(b, a, 2)); // backward}else if(dir == 'u'){edges[a].push_back(Edge(a, b, 2));edges[b].push_back(Edge(b, a, 1));}else{edges[a].push_back(Edge(a, b, 0)); // it's a rooted tree, so we don't store edge to father}}}if(have_data) {for(int i = 1; i <= n; i++)if(!have_father[i] && !edges[i].empty()) { root = i; break; }}return have_data;
}struct UndirectedSon {int w, f, g;UndirectedSon(int w=0, int f=0, int g=0):w(w),f(f),g(g){}
};bool cmp_f(const UndirectedSon& w1, const UndirectedSon& w2) {return w1.f < w2.f;
}bool cmp_g(const UndirectedSon& w1, const UndirectedSon& w2) {return w1.g < w2.g;
}
// calculate f[i] and g[i]
// return true iff f[i] < INF
// f[i] is the minimal length of the longest "->u" path if all subtree paths have length <= maxlen
// g[i] is the minimal length of the longest "u->" path if all subtree paths have length <= maxlen
// f[i] = g[i] = INF if "all subtree paths have length <= maxlen" cannot be satisfied
bool dp(int i, int fa) {if(edges[i].empty()) {f[i] = g[i] = 0;return true;}vector<UndirectedSon> sons;int f0 = 0, g0 = 0; // f'[i] and g'[i] for directed sons// let f'[i] = max{f[w] | w->i}+1, g'[i] = max{g[w] | i->w}+1// then we should change some undirected edges to ->u or u-> edges so that f'[i]+g'[i] <= maxlen// then f[i] is the minimal f'[i] under this condition, and g[i] is the minimal g'[i]for(int k = 0; k < edges[i].size(); k++) {int w = edges[i][k].v;if(w == fa) continue;dp(w, i);int d = edges[i][k].d;if(d == 0) sons.push_back(UndirectedSon(w, f[w], g[w]));else if(d == 1) g0 = max(g0, g[w]+1);else f0 = max(f0, f[w]+1);}// If there is no undirected edges, we're doneif(sons.empty()) {f[i] = f0; g[i] = g0;if(f[i] + g[i] > maxlen) { f[i] = g[i] = INF; }return f[i] < INF;}f[i] = g[i] = INF;// to calculate f[i], we sort f[w] of undirected sons in increasing order and make first p edges to w->i// then we calculate f'[i] and g'[i], check for f'[i]+g'[i] <= maxlen and update answerint s = sons.size();sort(sons.begin(), sons.end(), cmp_f);int maxg[maxn]; // maxg[i] is max{sons[i].g, sons[i+1].g, ...}maxg[s-1] = sons[s-1].g;for(int k = s-2; k >= 0; k--)maxg[k] = max(sons[k].g, maxg[k+1]);for(int p = 0; p <= sons.size(); p++) {int ff = f0, gg = g0;if(p > 0) ff = max(ff, sons[p-1].f+1);if(p < sons.size()) gg = max(gg, maxg[p]+1);if(ff + gg <= maxlen) f[i] = min(f[i], ff);}// g[i] is similarsort(sons.begin(), sons.end(), cmp_g);int maxf[maxn]; // maxf[i] is max{sons[i].f, sons[i+1].f, ...}maxf[s-1] = sons[s-1].f;for(int k = s-2; k >= 0; k--)maxf[k] = max(sons[k].f, maxf[k+1]);for(int p = 0; p <= sons.size(); p++) {int ff = f0, gg = g0;if(p > 0) gg = max(gg, sons[p-1].g+1);if(p < sons.size()) ff = max(ff, maxf[p]+1);if(ff + gg <= maxlen) g[i] = min(g[i], gg);}return f[i] < INF;
}int main() {while(read_data()) {maxlen = 0;for(int i = 1; i <= n; i++) maxlen = max(maxlen, dfs(i));// Note: the problem asks for the number of nodes in path, but all the "lengths" above mean "number of edges"if(dp(root, -1)) cout << maxlen+1 << "\n";else cout << maxlen+2 << "\n";}return 0;
}
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