CF708D Incorrect Flow
https://www.luogu.com.cn/problem/CF708D
分两种情况讨论
-
f
<
=
c
f <= c
f<=c
可以把 f f f减小,退流连边 ( v , u , 0 , f , 1 ) (v,u,0,f,1) (v,u,0,f,1)
可以把 f f f增大,流连边 ( u , v , 0 , c − f , 1 ) (u,v,0,c-f,1) (u,v,0,c−f,1)
可以将两个同时增大,连边 ( u , v , 0 , i n f , 2 ) (u,v,0,inf, 2) (u,v,0,inf,2)
强制先流 f f f,连边 ( u , v , f , f , 0 ) (u,v,f,f,0) (u,v,f,f,0) -
f
>
c
f > c
f>c
可以将两个同时增大,连边 ( u , v , 0 , i n f , 2 ) (u,v,0,inf, 2) (u,v,0,inf,2)
强制先流 f f f,连边 ( u , v , f , f , 0 ) (u,v,f,f,0) (u,v,f,f,0)
首先至少要花费 f − c f-c f−c使得 f = c f=c f=c,先计算
然后连一条推流边 ( v , u , 0 , f − c , 0 ) (v,u,0,f-c,0) (v,u,0,f−c,0)表示让 f = c f=c f=c在原来那个区间里移动
连一条反边 ( v , u , 0 , c , 0 ) (v,u,0,c,0) (v,u,0,c,0)表示将 f f f减小
然后跑上下界网络流即可
code:
#include<bits/stdc++.h>
#define N 805
#define ll long long
using namespace std;
struct edge {
int v, c, nxt;
ll w;
} e[N * N << 1];
int p[N], eid;
void init() {
memset(p, -1, sizeof p);
eid = 0;
}
void insert(int u, int v, int c, ll w) {
e[eid].v = v;
e[eid].c = c;
e[eid].w = w;
e[eid].nxt = p[u];
p[u] = eid ++;
}
void add(int u, int v, int c, ll w) { //printf("%d --> %d %d %d\n", u, v, c, w);
insert(u, v, c, w), insert(v, u, 0, -w);
}
int n, m, S, T, pre[N], vis[N];
ll dis[N];
queue<int> q;
const ll INF = 1e18;
const int inf = 1e9;
int bfs() {
for(int i = 0; i <= T; i ++) dis[i] = INF, vis[i] = 0, pre[i] = -1;
dis[S] = 0; q.push(S);
while(q.size()) {
int u = q.front(); q.pop();
vis[u] = 0;
for(int i = p[u]; i + 1; i = e[i].nxt) {
int v = e[i].v; //printf("*%d %d %d %d %d %d\n", u, v, dis[u], dis[v], e[i].c, e[i].w);
if(e[i].c && dis[u] + e[i].w < dis[v]) {
dis[v] = dis[u] + e[i].w;
pre[v] = i;
if(!vis[v]) vis[v] = 1, q.push(v);
}
}
}
// for(int i = 1; i <= T; i ++) printf("%lld ", dis[i]); printf("\n");
return pre[T] != -1;
}
ll mcfc() {
ll ret = 0;
for(; bfs() ;) {
ll flow = INF;
for(int i = T; i != S; i = e[pre[i] ^ 1].v) flow = min(flow, (ll)e[pre[i]].c);
for(int i = T; i != S; i = e[pre[i] ^ 1].v) {
ret += flow * e[pre[i]].w;
e[pre[i]].c -= flow, e[pre[i] ^ 1].c += flow;
}
// printf("*%d*", flow);
// break;
}
return ret;
}
ll sum[N], totc;
void addl(int u, int v, int l, int r, int c) { //printf("%d-->%d %d %d %d\n", u, v, l, r, c);
if(r - l > 0) add(u, v, r - l, c);
sum[u] -= l, sum[v] += l; //totc += 1ll * c * l;
}
int SS, TT;
void build() {
for(int i = 1; i <= TT; i ++) {
if(sum[i] > 0) add(S, i, sum[i], 0);
if(sum[i] < 0) add(i, T, - sum[i], 0);
}
// for(int i = 1; i <= TT; i ++) printf(" %lld ", sum[i]); printf("\n");
add(TT, SS, inf, 0);
}
int main() {
init();
scanf("%d%d", &n, &m);
SS = 1, TT = n; S = TT + 1, T = S + 1;
for(int i = 1; i <= m; i ++) {
int u, v, f, c;
scanf("%d%d%d%d", &u, &v, &c, &f);
if(f <= c) {
addl(v, u, 0, f, 1);
addl(u, v, 0, c - f, 1);
} else {
totc += f - c;
addl(v, u, 0, f - c, 0);
addl(v, u, 0, c, 1);
}
addl(u, v, f, f, 0);
addl(u, v, 0, inf, 2);
}
build();
printf("%lld", totc + mcfc());
return 0;
}