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);
}