Codeforces Gym 100814 G It is all about wisdom 二分+最短路

来源：互联网发布：银行存款怎样划算知乎编辑：程序博客网时间：2024/06/04 19:10

Foki is the president of the Martian United States of Altanie, Altanie is a very large and strange country. Each citizen in it has a positive integer wisdom value calculated based on her/his age and educational level (of course Foki has the maximum value). Altanie has a big map for all its roads, this map has the following properties:

There are N cities in Altanie, and the cities are numbered from 1 to N.
Each road connects 2 different cities, and all roads are bidirectional.
Each road requires a minimal wisdom value for the citizen to have the right to use it.
Each road costs some amount of Martian money to use it.
There is at most one road between each 2 cities.

Foki cares about all people of his country, so he is wondered about the minimum wisdom value that is needed to go from city 1 to city N with a total cost less than K, your job is to answer this question for him.

Input

The first line of the input contains T the number of the test cases. The first line of each test contains 1 < N ≤ 105 the number of the cities in Altanie, 1 ≤ M ≤ 105, the number of roads connecting the N cities and 1 ≤ K ≤ 109 the total cost.

Each of the next M lines contain a description of one of the M roads, each road is described with 1 ≤ s1, s2 ≤ N the numbers of two cities the road connects,1 ≤ c ≤ 109 the cost you have to pay each time you use this road, 1 ≤ W ≤ 109the minimal amount of wisdom value needed to have the right to use the road.

Output

For each test case print one line contains the answer of the following question: What is the minimum wisdom value a citizen should have to be able to go from city 1 to city N with cost less than K? if there is no solution, print -1.

Examples
input
25 6 31 2 1 11 4 1 11 3 1 12 5 2 12 4 1 13 5 1 55 6 21 2 1 11 4 1 11 3 1 12 5 2 12 4 1 13 5 1 5
output
5-1

Note

Warning: large Input/Output data, be careful with certain languages.

一个无向图，没有重边。所有路有一个花费，有一个最小智慧值，经过某条路径需要本人的智慧值大于等于所有路径的值智慧。问花费不超过k时，从1到n的最小智慧值是多少。

时限11s，放的这么开应该想到二分答案。

二分最大智慧值，每次用spfa寻找有没有这样一条路径。

#include <cstdio>#include <iostream>#include <string.h>#include <string> #include <map>#include <queue>#include <vector>#include <set>#include <algorithm>#include <math.h>#include <cmath>#include <bitset>#define mem0(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,0x3f,sizeof(a))using namespace std;typedef long long ll;typedef long double ld;const int maxn=100005,inf=0x3f3f3f3f;const ll llinf=0x3f3f3f3f3f3f3f3f; const ld pi=acos(-1.0L);int head[maxn],dist[maxn];int num,k,n,m;bool inque[maxn];struct Edge {int from,to,pre,dist,wis;};Edge edge[maxn*2];void addedge (int from,int to,int dist,int wis) {edge[num]=(Edge){from,to,head[from],dist,wis};head[from]=num++;edge[num]=(Edge){to,from,head[to],dist,wis};head[to]=num++;}bool find(int n,int maxd,int mwis) {queue<int> q;q.push(1);meminf(dist);mem0(inque);dist[1]=0;inque[1]=1;while (!q.empty()) {int now=q.front();q.pop();inque[now]=0;for (int i=head[now];i!=-1;i=edge[i].pre) {int to=edge[i].to;if (dist[now]+edge[i].dist<dist[to]&&dist[now]+edge[i].dist<k&&edge[i].wis<=mwis) {dist[to]=dist[now]+edge[i].dist;if (to==n) return true;if (!inque[to]) {inque[to]=1;q.push(to);}}}}return false;}int solve(int n) {int l,r,mid,ans=-1;l=1;r=1e9;while (l<=r) {mid=(l+r)/2;if (find(n,k,mid)) {ans=mid;r=mid-1;} else l=mid+1;}return ans;}int main() {int cas;scanf("%d",&cas);while (cas--) {num=0;scanf("%d%d%d",&n,&m,&k);int i,j,x,y,w,d;memset(head,-1,sizeof(head));for (i=1;i<=m;i++) {scanf("%d%d%d%d",&x,&y,&d,&w);addedge(x,y,d,w);}int ans=solve(n);printf("%d\n",ans);}return 0;}

阅读全文

0 0