题目

传送门

题解

建立s和t，然后s向1连下限0上限inf费用0的边，除1外所有节点向t连下限0上限inf费用0的边，对于每条边下限为1上限为inf费用为经过费用，然后我们只有做上下界网络流构出新图，跑最小费用可行流即可。

至于建立新图，我是这样建立的（如图）
另外推荐一篇文章：Menci的博客

代码

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int inf = INT_MAX;
const int maxn = 305 * 4;
const int M = 10000;
struct edge1 {
int from;
int to;
int low;
int high;
int cost;
};
vector<edge1> es;
struct edge {
int from;
int to;
int cap;
int cost;
};
vector<edge> edges;
vector<int> G[maxn];
inline void add_edge(int from, int to, int cap, int cost) {
edges.push_back((edge){from, to, cap, cost});
edges.push_back((edge){to, from, 0, -cost});
int m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
inline int read() {
char c = getchar();
int f = 1, x = 0;
while (!isdigit(c)) {
if (c == '-')
f = -1;
c = getchar();
}
while (isdigit(c))
x = x * 10 + c - '0', c = getchar();
return x * f;
}
int n, s, t, V;
void build_network() {
for (int i = 0; i < es.size(); i++) {
edge1 &e = es[i];
add_edge(e.from, e.to, e.high - e.low, e.cost);
add_edge(e.from, e.to, e.low, e.cost - M);
}
}
int dist[maxn], a[maxn], pree[maxn], inq[maxn];
bool spfa(int s, int t, int &cost) {
for (int i = 0; i < V; i++)
dist[i] = inf;
memset(pree, 0, sizeof(pree));
memset(inq, 0, sizeof(inq));
a[s] = inf;
dist[s] = 0;
queue<int> q;
q.push(s);
inq[s] = 1;
while (!q.empty()) {
int u = q.front();
q.pop();
inq[u] = 0;
for (int i = 0; i < G[u].size(); i++) {
edge &e = edges[G[u][i]];
if (e.cap > 0 && dist[e.to] > dist[u] + e.cost) {
pree[e.to] = G[u][i];
dist[e.to] = dist[u] + e.cost;
a[e.to] = min(e.cap, a[u]);
if (!inq[e.to]) {
q.push(e.to);
inq[e.to] = 1;
}
}
}
}
if (dist[t] >= inf)
return false;
cost += a[t] * dist[t];
int u = t;
while (u != s) {
edges[pree[u]].cap -= a[t];
edges[pree[u] ^ 1].cap += a[t];
u = edges[pree[u]].from;
}
return true;
}
int mcmf(int s, int t) {
int cost = 0;
while (spfa(s, t, cost))
;
//    cout << "Hey:" << cost << endl;
return cost;
}
int main() {
//  freopen("input", "r", stdin);
scanf("%d", &n);
s = 0, t = n + 1, V = t + 1;
int cnt = 0;
es.push_back((edge1){s, 1, 0, inf, 0});
for (int i = 1; i <= n; i++) {
int k;
k = read();
for (int j = 0; j < k; j++) {
int a, b;
a = read();
b = read();
es.push_back((edge1){i, a, 1, inf, b});
cnt += 1;
}

es.push_back((edge1){i, t, 0, inf, 0});
}
build_network();
int ans = mcmf(s, t);
printf("%d", ans + M * cnt);
}