The 2019 Asia Nanchang First Round Online Programming Contest B Fire-Fighting Hero(阅读理解）
This is an era of team success, but also an era of heroes. Throughout the ages, there have been numerous examples of using the few to defeat the many. There are VV (Numbers 11 to VV) fire-fighting points in ACM city. These fire-fighting points have EE roads to communicate with each other. Among them, there is a fire-fighting hero in the SS fire-fighting point, and the fire-fighting team is distributed in K fire-fighting points. If a fire-fighting point needs to be put out, the fire-fighting hero or the fire-fighting team must arrive as soon as possible, that is, to choose the shortest route to arrive.
Today, our fire-fighting heroes want to challenge the fire-fighting team. The challenge is to: The maximum value of the shortest path for a fire-fighting hero to go to others fire-fighting points is compared with the maximum value of the shortest path for a fire-fighting team to go to others fire-fighting points from any point in their fire-fighting points. Because firefighting heroes are different and run faster, the maximum value of the shortest path they get should be discounted first, that is, multiplied by a coefficient of \frac{1}{C}C1, and then compared. The smaller one wins. Who is the real firefighter in this situation?
Who is the real firefighter in this situation?
Input
The first line contains a positive integer T (1\le T \le 10)T(1≤T≤10), which indicates that there are TT cases of test data.
The format of each case of test data is as follows:
- Line 11 contains five positive integers V (1 \le V \le 1000)V(1≤V≤1000), E (V-1 \le E \le \frac{V*V}{2})E(V−1≤E≤2V∗V), S (1 \le S \le V)S(1≤S≤V), K (1\le K \le V)K(1≤K≤V) and C (1\le C\le 10)C(1≤C≤10), the meanings are shown above.
- Line 22 contains KK positive integers, which in turn denotes the location number of the fire-fighting point where the fire-fighting team is located.
In the next EE line, three positive integers i, j (1 \le i, j \le V)i,j(1≤i,j≤V) and L (1 \le L \le 10000)L(1≤L≤10000) per line. Represents a path, i, ji,j as the endpoint (fire-fighting point), LL as the length of the path.
Output
Each case of test data outputs one line, which is a integer. That is, the maximum value of the shortest path of the winner (If the fire hero wins, the maximum value before the discount should be output). A draw is also a victory for fire-fighting hero.
样例输入复制
1
4 7 3 2 2
1 4
1 2 7
1 3 2
1 4 6
2 1 1
2 4 1
3 2 1
3 4 3样例输出复制
2
这个题当时交了89遍floyd，回去写了一遍超级源点，过了，这个垃圾题，是一个裸的超级源点，妈蛋，我就是读不懂他在说什么，这道题就是考察读题意。差评。
#include <cstdio>
#include <string>
#include <cmath>
#include <queue>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cstring>
#define ll long long
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
struct Node
{
int to, lat, val;
};
Node edge[1000005];
int head[1005],tot,dis[1005],N;
void add(int a, int b, int c)
{
edge[++tot].to = b;
edge[tot].val = c;
edge[tot].lat = head[a];
head[a] = tot;
}
void dijkstra(int yy)
{
priority_queue<pair<int,int>> pq;
for (int i = 1; i <= N; ++i)
dis[i] = inf;
dis[yy] = 0;
pq.push(make_pair(0,yy));
while (!pq.empty())
{
int s=pq.top().second;
pq.pop();
for (int i = head[s]; i; i = edge[i].lat)
{
int val = edge[i].to;
if (dis[val] > dis[s] + edge[i].val)
{
dis[val] = dis[s] + edge[i].val;
int w= dis[s] + edge[i].val;
pq.push(make_pair(-w,val));
}
}
}
}
int main()
{
int val, e, s, k, c;
int t, x;
scanf("%d", &t);
while (t--)
{
memset(head, 0, sizeof(head));
tot = 0;
scanf("%d %d %d %d %d", &N, &e, &s, &k, &c);
for (int i = 1; i <= k; ++i)
{
scanf("%d", &x);
add(0, x, 0);
add(x, 0, 0);
}
while (e--)
{
int a, b, dis;
scanf("%d %d %d", &a, &b, &dis);
add(a, b, dis),add(b,a,dis);
}
dijkstra(0);
int ans1 = -1;
for (int i = 1; i <= N; ++i) ans1 = max(ans1, dis[i]);
dijkstra(s);
int ans2= -1;
for (int i = 1; i <= N; ++i) ans2 = max(ans2, dis[i]);
if (ans1 * c < ans2)
printf("%d\n", ans1);
else
printf("%d\n", ans2);
}
return 0;
}
这题也没构造啥菊花图，这图那图的，用SPFA照样过，我真是个弟弟，89遍，我太垃圾了。
#include<iostream>
#include<queue>
#include<algorithm>
#include<set>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<bitset>
#include<cstdio>
#include<cstring>
#define Swap(a,b) a^=b^=a^=b
#define cini(n) scanf("%d",&n)
#define cinl(n) scanf("%lld",&n)
#define cinc(n) scanf("%c",&n)
#define cins(s) scanf("%s",s)
#define coui(n) printf("%d",n)
#define couc(n) printf("%c",n)
#define coul(n) printf("%lld",n)
#define speed ios_base::sync_with_stdio(0)
#define Max(a,b) a>b?a:b
#define Min(a,b) a<b?a:b
#define mem(n,x) memset(n,x,sizeof(n))
#define INF 0x3f3f3f3f
#define maxn 100010
#define Ege 100000000
#define Vertex 1005
#define esp 1e-9
#define mp(a,b) make_pair(a,b)
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
struct Node
{
int to, lat, val; //边的右端点，边下一条边，边权
};
Node edge[1000005];
int head[1005],tot,dis[1005],N,M,sb[1005],vis[1005];
void add(int from, int to, int dis)
{
edge[++tot].lat = head[from];
edge[tot].to = to;
edge[tot].val = dis;
head[from] = tot;
}
void spfa(int s)
{
memset(dis, 0x3f, sizeof(dis));
dis[0]=0;
memset(vis, 0, sizeof(vis));
vis[s] = 1;
dis[s] = 0;
queue<int>Q;
Q.push(s);
while (!Q.empty()) {
int u = Q.front();
Q.pop();
vis[u] = 0;
for (int i = head[u];i;i = edge[i].lat) {
int to = edge[i].to;
int di = edge[i].val;
if (dis[to]>dis[u] + di) {
dis[to] = dis[u] + di;
if (!vis[to]) {
vis[to] = 1;
Q.push(to);
}
}
}
}
}
int main()
{
int val, e, s, k, c;
int t, x;
scanf("%d", &t);
while (t--)
{
memset(head, 0, sizeof(head));
tot = 0;
scanf("%d %d %d %d %d", &N, &e, &s, &k, &c);
for (int i = 1; i <= k; ++i)
{
scanf("%d", &x);
add(0, x, 0);
add(x, 0, 0);
}
while (e--)
{
int a, b, dis;
scanf("%d %d %d", &a, &b, &dis);
add(a, b, dis),add(b,a,dis);
}
spfa(0);
int ans1 = -1;
for (int i = 1; i <= N; ++i) ans1 = max(ans1, dis[i]);
spfa(s);
int ans2= -1;
for (int i = 1; i <= N; ++i) ans2 = max(ans2, dis[i]);
if (ans1 * c < ans2)
printf("%d\n", ans1);
else
printf("%d\n", ans2);
}
return 0;
}
