HDU 4406 - GPA(网络流‘费用流)

题目：

http://acm.hdu.edu.cn/showproblem.php?pid=4406

题意：

复习m门课，有n天的复习时间，每一天只能复习k个课时。给出m门课的学分和基础分。n*m的矩阵：1表示i天可以复习课程j 。花一个课时复习该课程则可以增加一分。

在使得没有课程不及格的情况下，求出最高的GPA（给出相关公式）。

思路：

每科的绩点f(x, w) = (4.0 - 3.0 * (100 - x) * (100 - x) / 1600) * w, 可以看出是 费用 与流量的平方成正比的费用流，所以使用 拆边法 （白书）。

建边：

1：分数< 60: 建一条容量为60-基础分，费用为-inf 的边，保证该课程先满足60分。对于分数（61,100），对于每个分数建一条容量为1，费用为 - （f（x，w）- f（x-1，w) ).

  若分数>= 60，则按照第二建边方法建边。

2：每门课到每一天建 容量为K，费用0的边。

3：每一天到汇点建 容量为K，费用为0的边。

AC.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>

using namespace std;
const double inf = 0x3f3f3f3f;
const int MAXN = 1000;
const int MAXM = 20000;
int n, K, m;

int cre[25], baf[25], bf[25];

struct Edge
{
int to, next, cap, flow;
double cost;
} edge[MAXM];

int head[MAXN], tol;
int pre[MAXN];
double dis[MAXN];
bool vis[MAXN];
int N;

void init()
{
N = n+m+2;
tol = 0;
memset(head, -1, sizeof(head));
}

void addedge(int u, int v, int cap, double cost)
{
edge[tol].to = v;
edge[tol].cap = cap;
edge[tol].cost = cost;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = 0;
edge[tol].cost = -cost;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}

bool spfa(int s, int t)
{
queue<int>q;
for (int i = 0; i < N; i++)
{
dis[i] = inf;
vis[i] = false;
pre[i] = -1;
}
dis[s] = 0;
vis[s] = true;
q.push(s);
while (!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (edge[i].cap > edge[i].flow &&
dis[v] > dis[u] + edge[i].cost )
{
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if (!vis[v])
{
vis[v] = true;
q.push(v);
}
}
}
}
if (pre[t] == -1) return false;
else return true;
}

int minCostMaxflow(int s, int t, double &cost)
{
int flow = 0;
cost = 0;
while (spfa(s, t))
{
int Min = inf;
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
{
if (Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
}
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
{
edge[i].flow += Min;
edge[i ^ 1].flow -= Min;
// if (cost < edge[i].cost * Min + cost)
// return 0;
// else
cost += edge[i].cost * Min;
}
flow += Min;
}
return flow;
}

double cal(int x, int w)
{
return (4.0 - 3.0 * (100 - x) * (100 - x) / 1600) * w;
}
int main()
{
//freopen("in", "r", stdin);
while(~scanf("%d%d%d", &n, &K, &m)) {
if(n == 0 && K == 0 && m == 0) break;

init();
int Scre = 0;
for(int i = 1; i <= m; ++i) {
scanf("%d", &cre[i]);
Scre += cre[i];
}
for(int i = 1; i <= m; ++i) {
scanf("%d", &baf[i]);
}
int g;
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= m; ++j) {
scanf("%d", &g);
if(g == 0) continue;
addedge(j, i+m, K, 0);
}
}

int s = 0, t = n+m+1;
for(int i = 1; i <= n; ++i) {
addedge(i+m, t, K, 0);
}

double f1, f2;
for(int i = 1; i <= m; ++i) {
if(baf[i] < 60) {
addedge(s, i, 60-baf[i], -inf);
f2 = cal(60, cre[i]);
for(int j = 61; j <= 100; ++j) {
f1 = cal(j, cre[i]);
addedge(s, i, 1, -(f1-f2));
f2 = f1;

}
}
else {
f2 = cal(baf[i], cre[i]);
for(int j = baf[i]+1; j <= 100; ++j) {
f1 = cal(j, cre[i]);
addedge(s, i, 1, -(f1-f2));
f2 = f1;
}
}
}

double cost;
int fw = minCostMaxflow(s, t, cost);
//printf("%lf, %d\n", cost, fw);

bool flag = 1;
for(int i = head[s]; ~i; i = edge[i].next) {
int v = edge[i].to ;
if(edge[i].flow > 0) {
baf[v] += edge[i].flow;
}
}

double ans = 0;
for(int i = 1; i <= m; ++i) {
if(baf[i] < 60) {
flag = 0;
break;
}
ans += cal(baf[i], cre[i]) / Scre;
//printf("***%d\n", baf[i]);
}
if(!flag) {
printf("0.000000\n");
}
else printf("%.6lf\n", ans);
}
return 0;
}