SDUT 2414 An interesting game(最大费用流)

题目大意：给定初始的n个山坡，并且有m个可加的山坡，选其中的不超过k个山坡来，放入n个山坡中不能够（两两的中间）。问最后加入不超过k个山坡的高度差之和。

思路：将其中的n个山坡与原点联，n个山坡再与m个山坡高度联(0<=h<=30)直接哈希一下费用为abs(a[i]-j)+abs(a[i+1]-j)-abs(a[i]-a[i+1])，然后m个山坡的高度再与汇点前一个点联流量为该座山的使用次数，最后联汇点流量为k.

#include<map>
#include<queue>
#include<cmath>
#include<cstdio>
#include<stack>
#include<iostream>
#include<cstring>
#include<algorithm>
#define LL long long
#define inf 0x3f3f3f3f
#define eps 1e-8
#define ls l,mid,rt<<1
#define rs mid+1,r,rt<<1|1
const double PI=acos(-1.0);
using namespace std;

struct node{
int to,w,c,next;
}q[100000*5];
int head[1000000];
int st,ed,cnt,maxcost;
void add(int a,int b,int w,int c){
q[cnt].to = b;
q[cnt].w = w;
q[cnt].c = c;
q[cnt].next = head[a];
head[a] = cnt++;

q[cnt].to = a;
q[cnt].w = 0;
q[cnt].c = -c;
q[cnt].next = head[b];
head[b] = cnt++;
}

int ha[40],dis[2010],flow;
int a[2010],f[2010],cur[2010];
bool vis[2010];

bool SPFA(){

memset(dis,inf,sizeof(dis));
memset(vis,false,sizeof(vis));
memset(cur,-1,sizeof(cur));
queue<int>Q;
while(!Q.empty())
Q.pop();
Q.push(st);
f[st]=inf;
dis[st]=0;
vis[st] = true;
while(!Q.empty()){
int u = Q.front();
Q.pop();
vis[u] = false;
for(int i = head[u];~i;i=q[i].next){
int v = q[i].to;
if(dis[v] > dis[u] + q[i].c &&q[i].w > 0){
dis[v] = dis[u] + q[i].c;
cur[v] = i;
f[v] = min(f[u],q[i].w);
if(!vis[v]){
vis[v] = true;
Q.push(v);
}
}
}
}

return dis[ed] != inf;
}
int so(){
int ant=0;
for(int i = ed;i != st;i = q[cur[i]^1 ].to){
q[cur[i] ].w -= 1;
q[cur[i]^1 ].w +=1;
ant += q[cur[i] ].c;
}
return ant;
}
int main(){
int n,m,i,j,k,cla;
scanf("%d",&cla);
for(int zu = 1;zu <= cla;++ zu){
memset(ha,0,sizeof(ha));
maxcost=flow=cnt=st=0;
int ans=0;
memset(head,-1,sizeof(head));
memset(vis,false,sizeof(vis));
scanf("%d%d%d",&n,&m,&k);
for(i = 1;i <= n;++ i ){
scanf("%d",&a[i]);
}
for(i = 1;i <= m;++ i){
scanf("%d",&j);
++ ha[j];
}
for(i = 1 ;i < n;++ i){
add(st,i,1,0);
}
for(i = 1;i < n;i ++){
ans+=abs(a[i]-a[i+1]);
for(j = 0 ;j <= 30;++ j)
if(ha[j])
add(i,n + j,1,-(abs(a[i]-j)+abs(a[i+1]-j)-abs(a[i+1]-a[i])));
}
ed = n + 41;
for( i = 0;i <= 30 ;++ i )
if(ha[i])
add(i + n,ed-1,ha[i],0);
add(ed-1,ed,k,0);
int tmp=0;
while(SPFA()){
tmp += so();
}
printf("Case %d: %d\n",zu,ans-tmp);
}
return 0;
}