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
", 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
", (int)res.size());
for (int i = 0; i < (int)res.size(); i ++){
printf("%d %c
", res[i].first, res[i].second);
}
}
return 0;
}
[/cpp]