Holy Grail
限制
1000 ms
256 MB
As the current heir of a wizarding family with a long history,unfortunately, you fifind yourself
forced to participate in the cruel Holy Grail War which has a reincarnation of sixty
years.However,fortunately,you summoned a Caster Servant with a powerful Noble
Phantasm.When your servant launch her Noble Phantasm,it will construct a magic
fifield,which is actually a directed graph consisting of n vertices and m edges.More
specififically,the graph satisfifies the following restrictions :
Does not have multiple edges(for each pair of vertices x and y, there is at most one
edge between this pair of vertices in the graph) and does not have self-loops(edges
connecting the vertex with itself).
May have negative-weighted edges.
Does not have a negative-weighted loop.
n<=300 , m<=500.
Currently,as your servant's Master,as long as you add extra 6 edges to the graph,you will
beat the other 6 masters to win the Holy Grail.
However，you are subject to the following restrictions when you add the edges to the
graph:
Each time you add an edge whose cost is c,it will cost you c units of Magic
Value.Therefore,you need to add an edge which has the lowest weight(it's probably
that you need to add an edge which has a negative weight).
Each time you add an edge to the graph,the graph must not have negative
loops,otherwise you will be engulfed by the Holy Grail you summon.
Input
Input data contains multiple test cases. The fifirst line of input contains integer t — the
number of test
cases (1 ≤ t ≤ 5).
For each test case,the fifirst line contains two integers n,m,the number of vertices in the
graph, the initial number of edges in the graph.
Then m lines follow, each line contains three integers x, y and w (0 ≤ x, y < n,−109≤w≤
109, x = y) denoting an edge from vertices x to y (0-indexed) of weight w.Then 6 lines follow, each line contains two integers s,t denoting the starting vertex and
the ending vertex of the edge you need to add to the graph.
It is guaranteed that there is not an edge starting from s to t before you add any edges and
there must exists such an edge which has the lowest weight and satisfifies the above
restrictions, meaning the solution absolutely exists for each query.
Output
For each test case,output 6 lines.
Each line contains the weight of the edge you add to the graph.
Sample Input
1
10 15
4 7 10
7 6 3
5 3 3
1 4 11
0 6 20
9 8 25
3 0 9
1 2 15
9 0 27
5 2 0
7 3 -5
1 7 21
5 0 1
9 3 16
1 8 4
4 1
0 3
6 9
2 1
8 7
0 4
Sample Output-11
-9
-45
-15
17
7
比赛时用的SPFA,其实用Floyd也可以做，泪奔，awsl。。。。。。。。。。。。。。。。
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn=3e2+7,maxm=5e2+5; 5 const ll inf=1e16; 6 int head[maxm],vis[maxn],cnt=0; 7 ll dist[maxn]; 8 struct node 9 { 10 int to,next; 11 ll w; 12 } edge[maxm]; 13 void add(int a,int b,ll w) 14 { 15 edge[++cnt].to=b,edge[cnt].next=head[a]; 16 edge[cnt].w=w,head[a]=cnt; 17 } 18 queue<int>que; 19 void spfa(int t) 20 { 21 dist[t]=0,vis[t]=1; 22 que.push(t); 23 while(que.size()) 24 { 25 int now=que.front(); 26 que.pop(); 27 vis[now]=0; 28 for(int i=head[now]; i; i=edge[i].next) 29 { 30 int to=edge[i].to,w=edge[i].w; 31 if(dist[to]>dist[now]+w) 32 { 33 dist[to]=dist[now]+w; 34 if(!vis[to]) 35 { 36 que.push(to); 37 vis[to]=1; 38 } 39 } 40 } 41 } 42 } 43 int main() 44 { 45 int t; 46 scanf("%d",&t); 47 while(t--) 48 { 49 cnt=0; 50 memset(vis,0,sizeof(vis)); 51 memset(head,0,sizeof(head)); 52 for(int i=0; i<=maxn-2; ++i) 53 dist[i]=inf; 54 int n,m,a,b,s,t; 55 ll w; 56 scanf("%d%d",&n,&m); 57 for(int i=1; i<=m; ++i) 58 { 59 scanf("%d %d %lld",&a,&b,&w); 60 add(a,b,w); 61 } 62 for(int i=1; i<=6; ++i) 63 { 64 scanf("%d %d",&s,&t); 65 spfa(t); 66 printf("%lld ",-dist[s]); 67 add(s,t,-dist[s]); 68 memset(vis,0,sizeof(vis)); 69 for(int i=0; i<=maxn-2; ++i) 70 dist[i]=inf; 71 } 72 } 73 return 0; 74 }