思路分析
先求出容量为1的时候的最小费用最大流,记录每一次增广路的权值,对于询问只需要看需要几条增广路,前缀和处理一下就行了
AC代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
bool IOerror=0;
inline char nc() {
static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
if(p1==pend) {
p1=buf;
pend=buf+fread(buf,1,BUF_SIZE,stdin);
if(pend==p1) {
IOerror=1;
return -1;
}
}
return *p1++;
}
inline bool blank(char ch) {
return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
}
template<class T> inline bool read(T &x) {
bool sign=0;
char ch=nc();
x=0;
for(; blank(ch); ch=nc());
if(IOerror)return false;
if(ch=='-')sign=1,ch=nc();
for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
if(sign)x=-x;
return true;
}
template<class T,class... U>bool read(T& h,U&... t) {
return read(h)&&read(t...);
}
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = 0x3f3f3f3f;
const int N = 55;
const int M = 333;
#define int long long
struct node {
int v,flow,w,next;
} e[M];
int head[N],cnt;
int n,m,s,t,k,maxflow,mincost,path[N],pre[N];
int dis[N],vis[N],ss[333];
vector<int>ans;
inline void ade(int u,int v,int f,int w) {
e[++cnt].v=v;
e[cnt].flow=f;
e[cnt].w=w;
e[cnt].next=head[u];
head[u]=cnt;
}
inline void add(int u,int v,int f,int w) {
ade(u,v,f,w);
ade(v,u,0,-w);
}
inline bool spfa() {
mem(dis,INF);
mem(vis,0);
mem(pre,-1);
dis[s]=0;
vis[s]=1;
queue<int>q;
q.push(s);
while(!q.empty()) {
int u=q.front();
vis[u]=0;
q.pop();
for(int i=head[u]; i; i=e[i].next) {
int v=e[i].v;
int w=e[i].w;
if(e[i].flow && dis[v]>dis[u]+w) {
dis[v]=dis[u]+w;
pre[v]=u;
path[v]=i;
if(!vis[v]) {
q.push(v);
vis[v]=1;
}
}
}
}
return pre[t]!=-1;
}
inline void EK() {
while(spfa()) {
ans.pb(dis[t]);
int mi=INF;
for(int i=t; i!=s; i=pre[i])
mi=min(mi,e[path[i]].flow);
for(int i=t; i!=s; i=pre[i]) {
e[path[i]].flow-=mi;
e[path[i]^1].flow+=mi;
}
maxflow+=mi;
mincost+=dis[t]*mi;
}
}
signed main() {
while(read(n,m)){
s=1,t=n;
mem(head,0);
cnt=1;
ans.clear();
for(int i=1;i<=m;i++){
int u,v,w;
read(u,v,w);
add(u,v,1,w);
}
EK();
int q;
read(q);
int len=sz(ans);
for(int i=0;i<len;i++) ss[i+1]=ss[i]+ans[i];
while(q--){
int u,v;
read(u,v);
if(u*len<v) {
puts("NaN");
continue;
}
int k=v/u;
int w=ss[k];
int z=v-k*u;
w=w*u+ans[k]*z;
int g=__gcd(w,v);
printf("%lld/%lld\n",w/g,v/g);
}
}
return 0;
}