洛谷p3778[APIO2017]商旅

【洛谷p3778】【APIO2017】商旅

题面

洛谷

题解

01分数规划水题...
设总收益为\(x\)总代价为\(y\)，那么题目中求的就是\(\frac{x}{y}<=z\)的最小取值。
把这个式子变个形:\(x-yz<=0\)，然后二分\(z\)。
我们可以\(O(n^3)\)预处理出最大收益\(pro_{i,j}\)和这张图每2个点之间的最短耗时（其实就是最短路）\(d_{i,j}\)，
然后连边\(d_{i,j}*z-pro_{i,j}\)用\(spfa\)判一下负环即可。

代码

#include <bits/stdc++.h>

typedef long double db;  
const db eps = 1e-9;
const int maxn = 110;
const int maxk = 1010;
const int inf = 0x3f3f3f3f;   

template<class t> inline void read(t& res) {
	res = 0;  char ch = getchar();  bool neg = 0;
	while(!isdigit(ch))
		neg |= ch == '-', ch = getchar();
	while(isdigit(ch))
		res = (res << 1) + (res << 3) + (ch & 15), ch = getchar();
	if(neg)
		res = -res;  
}
inline void cmin(int& a,int b) {
	if(a > b)
		a = b;
}
inline void cmax(int& a,int b) {
	if(a < b)
		a = b;    
}

int S[maxk][maxk], B[maxk][maxk];  
int pro[maxk][maxk], d[maxn][maxn], cnt[maxn];  
int hd[maxn], ver[maxn * maxn], nxt[maxn * maxn];  db wei[maxn * maxn];    
db dis[maxn];  bool vis[maxn];  
int n, m, i, j, k, K, cnte;

inline bool equal(db a,db b) { return b - a < eps; }
inline void adde(int u,int v,db w) {
	ver[++cnte] = v;  wei[cnte] = w;   
	nxt[cnte] = hd[u];  hd[u] = cnte;  
}

inline bool check(db mid) {
	std::queue<int> q;
	memset(hd,-1,sizeof(hd));  cnte = 0;
	while(!q.empty())
		q.pop();
	for(int i = 1;i <= n;i++)
		for(int j = 1;j <= n;j++)
			if(d[i][j] != inf && i != j)
				adde(i,j,d[i][j] * mid - pro[i][j]);
	for(int i = 1;i <= n;i++)
		q.push(i), dis[i] = 0, vis[i] = 1, cnt[i] = 1;
	while(!q.empty()) {
		int u = q.front();  q.pop();  vis[u] = 0;     
		if(cnt[u] > n)
			return 1;
		for(int i = hd[u];~i;i = nxt[i]) {
			int v = ver[i];  db w = wei[i];   
			if(dis[u] + w < dis[v]) {
				dis[v] = dis[u] + w;  
				if(!vis[v]) {
					vis[v] = 1;
					cnt[v]++;
					q.push(v);
				}
			}
		}
	}
	return 0;      
}

int main() {
	read(n);  read(m);  read(K);
	for(int i = 1;i <= n;i++)
		for(int j = 1;j <= K;j++)
			read(B[i][j]), read(S[i][j]);  K++;    
	for(int i = 1;i <= n;i++)
		for(int j = 1;j <= n;j++)	
			for(int k = 1;k <= K;k++)
				if(~S[j][k] && ~B[i][k])
					cmax(pro[i][j],S[j][k] - B[i][k]);     
	memset(d,0x3f,sizeof(d));  
	for(int i = 1, u, v, w;i <= m;i++) {
		read(u);  read(v);  read(w);
		d[u][v] = w;   
	}
	for(int i = 1;i <= n;i++)
		d[i][i] = 0;
	for(int k = 1;k <= n;k++)
		for(int i = 1;i <= n;i++)
			for(int j = 1;j <= n;j++)
				cmin(d[i][j],d[i][k] + d[k][j]);  	
 	db l = 0, r = 1e9;		
	while(!equal(l,r)) {
		db mid = (l + r) / 2.0;   
		if(check(mid))
			l = mid;
		else
			r = mid;
	}
	printf("%d\n",(int)(floor(r)));
	return 0;  
}