luogu P3305 [SDOI2013]费用流

题目链接

bz似乎挂了...

luogu P3305 [SDOI2013]费用流

题解

dalao告诉我，这题

似乎很水....

懂了题目大意就可以随便切了

问1，最大流

问2，二分最大边权求，check是否为最大流

PS：今天代码很多bug....惨orz

代码

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define INF 0x7fffffff
#define eps 1e-4
const int maxn = 5007;
using std::queue;
using std::min;
int n,m,S,T;

inline int read() {
	int x = 0,f = 1;
	char c = getchar();
	while(c < '0' || c > '9') {if(f == '-') f = -1;c = getchar();}
	while(c <= '9'&&c >= '0') x = x * 10 + c - '0',c = getchar();
	return x*f;
}
struct node{
	int v;double flow;int next;
}edge[maxn<<2];
int head[maxn],num=1;
inline void add_edge(int u,int v,double flow) { edge[++num].v = v;edge[num].flow = flow;edge[num].next = head[u];head[u] = num; }
inline void ADD (int u,int v,double flow) {
	add_edge(u,v,flow);
	add_edge(v,u,0);
}
int a[maxn],b[maxn];double c[maxn];int lev[maxn],cur[maxn];
bool bfs() {
	for(int i = 0;i <= n;++ i) lev[i] = -1,cur[i] = head[i];
	queue<int>q;q.push(S),lev[S] = 0;
	while(!q.empty()) {
		int u = q.front();q.pop();
		for(int i = head[u];i;i = edge[i].next) {
			int v = edge[i].v;
			if(lev[v] == -1&&edge[i].flow > 0)
				lev[v] = lev[u]+1,q.push(v);
		}
	}
	if(lev[T] != -1)return true;
	return false;
}
double dfs(int x,double flow) {
	if(x == T)return flow;
	double delta,rest = 0;
	for(int v,&i = cur[x];i;i = edge[i].next) {
		v = edge[i].v;
		if(lev[v] == lev[x]+1&&edge[i].flow > 0) {
			delta = dfs(v,std::min(flow - rest,edge[i].flow));
			if(delta) {
				edge[i].flow -= delta;
				edge[i^1].flow += delta;
				rest+=delta;if(rest == delta)break;
			}
		}
	}
	if(rest==flow) lev[x]=-1;
	return rest;
}

double Dinic() {
	double ret = 0;
	while(bfs())
	ret += dfs(S,INF);
	return ret;
}
void rebuild(double mid) {
	memset(head,0,sizeof head);
	num=1;
	for(int i = 1;i <= m;++ i) {
		ADD(a[i],b[i],min(mid,c[i]));
	}
}
int main() {
	double p;
	n = read(),m = read(),scanf("%lf",&p);
	S=1;T=n;
	for (int i = 1;i <= m;++ i) scanf("%d%d%lf",&a[i],&b[i],&c[i]), ADD(a[i],b[i],c[i]);
	double Max_flow;
	Max_flow = Dinic();
	printf("%.0lf\n",Max_flow);
	double l = 0,r = 1000000086.0;
	while (r - l > eps) {
		double mid = (l + r) / 2;
		rebuild(mid);
		if(Dinic() != Max_flow) l = mid;
		else r = mid;
	}
	printf("%.4lf\n",l*p);
	return 0;
}