题意
有一个图， 两种操作，一种是删除某点的所有出边，一种是删除某点的所有入边，各个点的不同操作分别有一个花费，现在我们想把这个图的边都删除掉，需要的最小花费是多少。
思路
很明显的二分图最小点权覆盖集。WA在输出最小割方案上。 【
输出最小割方案】
从源点S做一次DFS遍历，标记所有访问到的点，这些点就是S点集。然后对于每一条满流边，如果其两端点一个在S点集一个不在则该边就是此方案下的最小割边。
代码
[cpp] #include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <algorithm> #include <string> #include <cstring> #define MID(x,y) ((x+y)/2) #define MEM(a,b) memset(a,b,sizeof(a)) #define REP(i, begin, m) for (int i = begin; i < begin+m; i ++) using namespace std; const int MAXV = 205; const int MAXE = 10005; const int oo = 0x3fffffff; template <typename T> struct Dinic{ struct flow_node{ int u, v; T flow; int opp; int next; }arc[2*MAXE]; int vn, en, head[MAXV]; int cur[MAXV]; int q[MAXV]; int path[2*MAXE], top; int dep[MAXV]; void init(int n){ vn = n; en = 0; MEM(head, -1); } void insert_flow(int u, int v, T flow){ arc[en].u = u; arc[en].v = v; arc[en].flow = flow; arc[en].next = head[u]; head[u] = en ++; arc[en].u = v; arc[en].v = u; arc[en].flow = 0; arc[en].next = head[v]; head[v] = en ++; } bool bfs(int s, int t){ MEM(dep, -1); int lq = 0, rq = 1; dep[s] = 0; q[lq] = s; while(lq < rq){ int u = q[lq ++]; if (u == t){ return true; } for (int i = head[u]; i != -1; i = arc[i].next){ int v = arc[i].v; if (dep[v] == -1 && arc[i].flow > 0){ dep[v] = dep[u] + 1; q[rq ++] = v; } } } return false; } T solve(int s, int t){ T maxflow = 0; while(bfs(s, t)){ int i, j; for (i = 1; i <= vn; i ++) cur[i] = head[i]; for (i = s, top = 0;;){ if (i == t){ int mink; T minflow = 0x7fffffff; //要比容量的oo大 for (int k = 0; k < top; k ++) if (minflow > arc[path[k]].flow){ minflow = arc[path[k]].flow; mink = k; } for (int k = 0; k < top; k ++) arc[path[k]].flow -= minflow, arc[path[k]^1].flow += minflow; maxflow += minflow; top = mink; i = arc[path[top]].u; } for (j = cur[i]; j != -1; cur[i] = j = arc[j].next){ int v = arc[j].v; if (arc[j].flow && dep[v] == dep[i] + 1) break; } if (j != -1){ path[top ++] = j; i = arc[j].v; } else{ if (top == 0) break; dep[i] = -1; i = arc[path[-- top]].u; } } } return maxflow; } }; Dinic <int> dinic; vector <pair<int, char> > res; bool vis[MAXV]; void dfs(int u){ vis[u] = true; for (int i = dinic.head[u]; i != -1; i = dinic.arc[i].next){ if (dinic.arc[i].flow <= 0) continue; int v = dinic.arc[i].v; if (!vis[v]){ dfs(v); } } } int main(){ //freopen("test.in", "r", stdin); //freopen("test.out", "w", stdout); int n, m; while(scanf("%d %d", &n, &m) != EOF){ dinic.init(2*n+2); REP(i, 1, n){ int tmp; scanf("%d", &tmp); dinic.insert_flow(i+n, 2*n+2, tmp); } REP(i, 1, n){ int tmp; scanf("%d", &tmp); dinic.insert_flow(2*n+1, i, tmp); } REP(i, 1, m){ int u, v; scanf("%d %d", &u, &v); dinic.insert_flow(u, v+n, oo); } printf("%d\n", dinic.solve(2*n+1, 2*n+2)); res.clear(); MEM(vis, 0); dfs(2*n+1); for (int i = 0; i < dinic.en; i ++){ if (dinic.arc[i].u == 2*n+1 && dinic.arc[i].flow == 0){ if (!vis[dinic.arc[i].v]) res.push_back(make_pair(dinic.arc[i].v, '-')); } if (dinic.arc[i].v == 2*n+2 && dinic.arc[i].flow == 0){ if (vis[dinic.arc[i].u]) res.push_back(make_pair(dinic.arc[i].u - n, '+')); } } printf("%d\n", (int)res.size()); for (int i = 0; i < (int)res.size(); i ++){ printf("%d %c\n", res[i].first, res[i].second); } } return 0; } [/cpp]
