题目大意:有N个工程,完成一个工程获利pi. M个技术问题,解决一个技术问题需要花费ci
现在给出每个工程需要解决的技术问题,和解决一个技术问题前还需要解决哪些技术问题
问最大获利是多少
解题思路:最大权闭合的题目,构图的话
正权的话肯定是完成一个工程的获利,负权的话就是解决每个问题的费用了
接着是连边,可以画出图,他所需要的前置条件其实就是它所指向的点,这样就可以得到一张有向图了,有了有向图了,就可以建图了
具体的理论详见胡伯涛:算法合集之《最小割模型在信息学竞赛中的应用》
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int MAXNODE = 210;
const int MAXEDGE = 100010;
typedef int Type;
const Type INF = 0x3f3f3f3f;
struct Edge {
int u, v, next;
Type cap, flow;
Edge() {}
Edge(int u, int v, Type cap, Type flow, int next): u(u), v(v), cap(cap), flow(flow), next(next) {}
};
struct ISAP {
int n, m, s, t;
Edge edges[MAXEDGE];
int head[MAXNODE], p[MAXNODE], num[MAXNODE], cur[MAXNODE], d[MAXNODE];
bool vis[MAXNODE];
void init(int n) {
this->n = n;
memset(head, -1, sizeof(head));
m = 0;
}
void AddEdge(int u, int v, Type cap) {
edges[m] = Edge(u, v, cap, 0, head[u]);
head[u] = m++;
edges[m] = Edge(v, u, 0, 0, head[v]);
head[v] = m++;
}
bool BFS() {
memset(vis, 0, sizeof(vis));
queue<int> Q;
for (int i = 0; i < n; i++)
d[i] = INF;
d[t] = 0;
vis[t] = 1;
Q.push(t);
while (!Q.empty()) {
int u = Q.front(); Q.pop();
for (int i = head[u]; ~i; i = edges[i].next) {
Edge &e = edges[i ^ 1];
if (!vis[e.u] && e.cap > e.flow) {
vis[e.u] = true;
d[e.u] = d[u] + 1;
Q.push(e.u);
}
}
}
return vis[s];
}
Type Augment() {
int u = t;
Type flow = INF;
while (u != s) {
Edge &e = edges[p[u]];
flow = min(flow, e.cap - e.flow);
u = edges[p[u]].u;
}
u = t;
while (u != s) {
edges[p[u]].flow += flow;
edges[p[u] ^ 1].flow -= flow;
u = edges[p[u]].u;
}
return flow;
}
Type Maxflow(int s, int t) {
this->s = s; this->t = t;
Type flow = 0;
BFS();
//如果s-->t走不通
if (d[s] >= n)
return 0;
memset(num, 0, sizeof(num));
for (int i = 0; i < n; i++)
cur[i] = head[i];
for (int i = 0; i < n; i++)
if (d[i] < INF)
num[d[i]]++;
int u = s;
while (d[s] < n) {
if (u == t) {
flow += Augment();
u = s;
}
bool ok = false;//纪录是否找到了下一个点
for (int i = cur[u]; ~i; i = edges[i].next) {
Edge &e = edges[i];
if (e.cap > e.flow && d[u] == d[e.v] + 1) {
ok = true;
p[e.v] = i;//点v由第i条边增广得到
cur[u] = i;//尝试到第i条边
u = e.v;
break;
}
}
//如果没找到下一个点,表示u到t的最短路要变长了,或者没路可走了
if (!ok) {
//找寻u到下一个点的最短路
int Min = n - 1;
for (int i = head[u]; ~i; i = edges[i].next) {
Edge &e = edges[i];
if (e.cap > e.flow)
Min = min(Min, d[e.v]);
}
if (--num[d[u]] == 0)//GAP优化
break;
num[d[u] = Min + 1]++;
cur[u] = head[u];
//返回前一个点,因为该点的最短距离已经变了
if (u != s)
u = edges[p[u]].u;
}
}
return flow;
}
}isap;
int n, m, source, sink;
int Sum, cas = 1;
void init() {
scanf("%d%d", &n, &m);
source = 0, sink = n + m + 1;
isap.init(sink + 1);
Sum = 0;
int c, k;
for (int i = 1; i <= n; i++) {
scanf("%d", &c);
isap.AddEdge(source, i, c);
Sum += c;
}
for (int j = 1; j <= m; j++) {
scanf("%d", &c);
isap.AddEdge(j + n, sink, c);
}
for (int i = 1; i <= n; i++) {
scanf("%d", &k);
for (int j = 1; j <= k; j++) {
scanf("%d", &c);
isap.AddEdge(i, n + c + 1, INF);
}
}
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= m; j++) {
scanf("%d", &c);
if (c) isap.AddEdge(n + i, n + j, INF);
}
}
int maxflow = isap.Maxflow(source, sink);
printf("Case #%d: %d\n", cas++, Sum - maxflow);
}
int main() {
int test;
scanf("%d", &test);
while (test--) {
init();
}
return 0;
}