【2019icpc南京站网络赛 - H】Holy Grail（最短路，spfa判负环）

题干：

As the current heir of a wizarding family with a long history,unfortunately, you find 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 field,which is actually a directed graph consisting of n vertices and m edges.More specifically,the graph satisfies 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 first line of input contains integer t — the number of testcases (1 \le t \le 51≤t≤5).

For each test case,the first 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 \le x,y<n0≤x,y<n,-10^9−109≤w≤10^9109, x \not = yx=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 satisfies the above restrictions, meaning the solution absolutely exists for each query.

Output

For each test case,output 66 lines.

Each line contains the weight of the edge you add to the graph.

样例输入复制

样例输出复制

-11
-9
-45
-15
17
7

题目大意：

给一张n个点m条边的带权有向图（权值可能为负），问你依次进行6个操作，每个操作给定起点和终点，让你在起点终点之间加一条边，问你这条边的边权最小是多少，可以保证新图中无负环。

解题报告：

二分权值判负环check就行了。然而标解好像不用二分，直接找到s到t的最短路就可以了。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 500 + 5;
const ll PYL = 1e12;
int n,m;
int tot,head[MAX];
struct Edge {int u,v;int ne;ll w;
} e[MAX*100];
void add(int u,int v,ll w) {e[++tot].u = u;e[tot].v = v;e[tot].w = w;e[tot].ne = head[u];head[u] = tot;
}
ll dis[MAX];
int cnt[MAX];
bool vis[MAX];
bool spfa(int st) {for(int i = 1; i<=n+1; i++) dis[i] = 1e12,vis[i] = 0,cnt[i] = 0;;queue<int> q;dis[st]=0;q.push(st);vis[st]=1; cnt[st]=1;while(q.size()) {int cur = q.front();q.pop();vis[cur]=0;for(int i = head[cur]; ~i; i = e[i].ne) {int v = e[i].v;if(dis[v] <= dis[cur] + e[i].w) continue;dis[v] = dis[cur] + e[i].w;if(!vis[v]) {cnt[v]++; vis[v]=1; q.push(v);if(cnt[v] > n) return 0;//}}}return 1;//没有负环返回1
}
int U[MAX],V[MAX],W[MAX];
int main()
{int T;cin>>T;while(T--) {scanf("%d%d",&n,&m);for(int i = 1; i<=m; i++) scanf("%d%d%d",&U[i],&V[i],&W[i]),U[i]++,V[i]++;for(int Q = 1; Q<=6; Q++) {ll l = 0,r = 2e12,ans,mid;int st,ed;scanf("%d%d",&st,&ed);st++,ed++;while(l<=r) {mid = (l+r)>>1;tot=0;memset(head,-1,sizeof head);for(int i = 1; i<=m; i++) add(U[i],V[i],W[i]);for(int i = 1; i<=n; i++) add(n+1,i,0);add(st,ed,mid-PYL);if(spfa(n+1)) ans = mid,r = mid-1;else l = mid+1;}U[++m] = st,V[m] = ed,W[m] = ans-PYL;printf("%lld\n",ans-PYL);}}return 0 ;
}