P2488 [SDOI2011]工作安排 费用流
作者:互联网
\(\color{#0066ff}{ 题目描述 }\)
你的任务是制定出一个产品的分配方案,使得订单条件被满足,并且所有员工的愤怒值之和最小。由于我们并不想使用Special Judge,也为了使选手有更多的时间研究其他两道题目,你只需要输出最小的愤怒值之和就可以了。
\(\color{#0066ff}{输入格式}\)
\(\color{#0066ff}{输出格式}\)
仅输出一个整数,表示最小的愤怒值之和。
\(\color{#0066ff}{输入样例}\)
2 3
2 2 2
1 1 0
0 0 1
1
2
1 10
1
2
1 6
\(\color{#0066ff}{输出样例}\)
24
\(\color{#0066ff}{数据范围与提示}\)
\(\color{#0066ff}{ 题解 }\)
显然是个费用流
考虑怎么建边
愤怒值对于每个员工完成工作的数量来分段
完成工作数量? 这不就是从员工流出去多少流吗
所以,从S向员工,连多条边,每条边的容量为每段的长
这样愤怒值的问题就解决了
注意,每个任务不止完成一次
所以员工向任务连容量为inf的边,任务向T连容量为需要次数的边
#include<bits/stdc++.h>
#define LL long long
LL in() {
char ch; LL x = 0, f = 1;
while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
return x * f;
}
const int maxn = 1e5;
const LL inf = 999999999999999LL;
struct node {
int to;
LL dis, can;
node *nxt, *rev;
node(int to = 0, LL dis = 0, LL can = 0, node *nxt = NULL)
: to(to), dis(dis), can(can), nxt(nxt) {}
void *operator new (size_t) {
static node *S = NULL, *T = NULL;
return (S == T) && (T = (S = new node[1024]) + 1024), S++;
}
};
LL dis[maxn], change[maxn];
node *head[maxn], *road[maxn];
bool vis[maxn];
bool f[505][505];
int n, s, t, m;
std::vector<LL> v[maxn], d[maxn];
bool spfa() {
for(int i = s; i <= t; i++) dis[i] = inf, change[i] = inf;
std::queue<int> q;
dis[s] = 0;
q.push(s);
while(!q.empty()) {
int tp = q.front(); q.pop();
vis[tp] = false;
for(node *i = head[tp]; i; i = i->nxt)
if(dis[i->to] > dis[tp] + i->dis && i->can) {
dis[i->to] = dis[tp] + i->dis;
change[i->to] = std::min(change[tp], i->can);
road[i->to] = i;
if(!vis[i->to]) vis[i->to] = true, q.push(i->to);
}
}
return change[t] != inf;
}
LL mcmf() {
LL cost = 0;
while(spfa()) {
cost += change[t] * dis[t];
for(int o = t; o != s; o = road[o]->rev->to) {
road[o]->can -= change[t];
road[o]->rev->can += change[t];
}
}
return cost;
}
void add(int from, int to, LL dis, LL can) {
head[from] = new node(to, dis, can, head[from]);
}
void link(int from, int to, LL dis, LL can) {
add(from, to, dis, can);
add(to, from, -dis, 0);
head[from]->rev = head[to];
head[to]->rev = head[from];
}
int main() {
m = in(), n = in();
t = m + n + 1, s = 0;
for(int i = m + 1; i <= m + n; i++) link(i, t, 0, in());
for(int i = 1; i <= m; i++)
for(int j = 1; j <= n; j++)
if(in()) link(i, m + j, 0, inf);
for(int i = 1; i <= m; i++) {
int k = in();
v[i].push_back(0);
for(int j = 1; j <= k; j++) v[i].push_back(in());
v[i].push_back(inf);
for(int j = 1; j <= k + 1; j++) d[i].push_back(in());
for(int j = 0; j <= k; j++) link(s, i, d[i][j], v[i][j + 1] - v[i][j]);
}
printf("%lld", mcmf());
return 0;
}
标签:费用,head,int,LL,0066ff,SDOI2011,P2488,color,dis 来源: https://www.cnblogs.com/olinr/p/10357255.html