BZOJ 2055: 80人环游世界 [上下界费用流]

时间：2023-03-08 15:33:59

2055: 80人环游世界

题意：n个点带权图，选出m条路径，每个点经过val[i]次，求最小花费

建图比较简单

s拆点限制流量m

一个点拆成两个，限制流量val[i]，需要用上下界

图中有边的连边，容量INF权值为花费

上下界最小费用流

那些容量为0的边不加也可以



该死我把费用流的加边打错了查了一会...

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

#define fir first

#define sec second

typedef long long ll;

const int N=1005, M=4e5+5, INF=1e9;

inline ll read(){

    char c=getchar();ll x=0,f=1;

    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}

    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}

    return x*f;

}

int n, m, s, s0, t, tot, extra[N], x;

struct edge{int v, c, f, w, ne, lower;}e[M];

int cnt=1, h[N];

inline void ins(int u, int v, int c, int w=0, int lower=0) {

	e[++cnt] = (edge){v, c, 0, w, h[u], lower}; h[u]=cnt;

	e[++cnt] = (edge){u, 0, 0, -w, h[v], lower}; h[v]=cnt;

}

int q[N], head, tail, d[N], inq[N];

pair<int, int> pre[N];

inline void lop(int &x) {if(x==N) x=1;}

bool spfa(int s, int t) {

	memset(inq, 0, sizeof(inq));

	memset(d, 127, sizeof(d));

	head=tail=1;

	q[tail++]=s; d[s]=0; inq[s]=1;

	pre[t].fir = -1;

	while(head!=tail) {

		int u=q[head++]; inq[u]=0; lop(head);

		for(int i=h[u];i;i=e[i].ne) {

			int v=e[i].v;

			if(e[i].c > e[i].f && d[v]>d[u]+e[i].w) {

				d[v]=d[u]+e[i].w;

				pre[v] = make_pair(u, i);

				if(!inq[v]) q[tail++]=v, inq[v]=1, lop(tail);

			}

		}

	}

	return pre[t].fir != -1;

}

int ek(int s, int t) {

	int flow=0, cost=0, x;

	while(spfa(s, t)) {

		int f=INF;

		for(int i=t; i!=s; i=pre[i].fir)

			f=min(f, e[pre[i].sec].c - e[pre[i].sec].f);

		flow+=f; cost+=f*d[t];

		//printf("f %d %d\n",f,d[t]);

		for(int i=t; i!=s; i=pre[i].fir)

			x=pre[i].sec, e[x].f+=f, e[x^1].f-=f;

	}

	//printf("ek %d %d\n",flow,cost);

	return cost;

}

int main() {

	freopen("in","r",stdin);

	n=read(); m=read(); s0=0; s=n+n+1; t=n+n+2;

	ins(s0, s, 0, 0, m);

	extra[s0]-=m, extra[s]+=m;

	for(int i=1; i<=n; i++) {

		x=read();

		ins(s, i, INF); ins(i+n, t, INF);

		ins(i, i+n, 0, 0, x);

		extra[i]-=x, extra[i+n]+=x;

	}

	for(int i=1; i<=n; i++)

		for(int j=i+1; j<=n; j++) {

			int val=read();

			if(val!=-1) ins(i+n, j, INF, val, 0);

		}

	int ss=t+1, tt=t+2, sum=0; tot=tt;

	for(int i=s0; i<=t; i++) {

		if(extra[i]>0) ins(ss, i, extra[i]), sum+=extra[i];

		if(extra[i]<0) ins(i, tt, -extra[i]);

	}

	ins(t, s0, INF);

	int ans=ek(ss, tt);

	printf("%d", ans);

}

相关文章