本文深入探讨了游戏开发领域的关键技术，包括游戏引擎、Unity、Cocos2d-X等，以及AI音视频处理的应用，如语义识别、物体检测、语音识别等。同时，文章还涉及了开发工具、大数据开发、测试等方面的内容，为读者提供了一个全面的技术视角。
Spring has passed by the land of Gensokyo, a reclusive realm in the far east, and the cherry trees have since shred their blossoms. However, the hanami(flower viewing) kept on going, with feasts being hosted day after day with no end in sight. Adding onto that, every time the feast is held, an unknown restless spiritual aura in Gensokyo also increases; however, nothing happens even while the spiritual aura rises, but when the spiritual aura rises, not a single person attempts to stop the feasts. As such, everyone who goes to the feast, be it human or youkai(phantoms), appears to be very suspicious. The truth is that Ibuki Suika, an oni(ogre) who loves drinks, feasts, and competition, uses her ability to control density to gather people to form a banquet. And she scatters herself to become mist to attend it.
There are N places and M roads in Gensokyo. Suika wants to made Q residents of Gensokyo have a feast at Hakurei Shrine as soon as possible. Each resident will start off from her residence, choose the fastest route and move at speed Vi. What's more, residents in Gensokyo prefer to "yukkuri shiteitte ne!"("take it easy!"), resident Ri will spend time Tij playing at place Pj once she gets there, excluding the starting and terminal. However, Suika can use her ability to increase one's Vi by Ai * sqrt(ci). As the limit of her ability, the total sum of ci cannot exceed a given number C.
What's the minimum time that Suika needs to wait for before all residents get to Hakurei Shrine to begin the feast.
Input
There are about 16 cases. Process to the end of file.
Each case begins with 3 integers 1 <= N <= 100, 1 <= M <= 1000, 1 <= Q <= 16. Then N places. Each place is identified by its name whose length is less than 30. Then M roads. Each road is represent by two string Pa, Pb, its two ends, and a real number Dab, the distance. Then Q residents. Each resident contains two strings Xi and Pi, her name and her residence, and 2 + N real numbers, Vi, Ai, Ti1, Ti2, ..., TiN. The case ends with a real number C. All real numbers are in range [0.1, 1000000].
Output
The minimum time on separate lines.
Answer having an absolute or relative error less than 1e-6 will be accepted. You can assume that the answer always exists.
Sample Input
2 2 1 Eientei Hakurei_Shrine Hakurei_Shrine Eientei 2.72 Eientei Hakurei_Shrine 3.14 Reisen_Udongein_Inaba Eientei 0.1 0.2 0.3 0.4 1 8 11 9 Hakurei_Shrine Forest_of_Magic(Marisa's_house) Forest_of_Magic(Alice's_house) Misty_Lake Scarlet_Devil_Mansion Mayohiga Poltergeist_Mansion Hakugyokurou Hakurei_Shrine Misty_Lake 218 Hakurei_Shrine Forest_of_Magic(Marisa's_house) 192 Hakurei_Shrine Forest_of_Magic(Alice's_house) 255 Misty_Lake Forest_of_Magic(Marisa's_house) 51 Misty_Lake Forest_of_Magic(Alice's_house) 49 Misty_Lake Mayohiga 66 Mayohiga Forest_of_Magic(Alice's_house) 77 Misty_Lake Scarlet_Devil_Mansion 221 Poltergeist_Mansion Forest_of_Magic(Alice's_house) 71 Poltergeist_Mansion Hakugyokurou 73 Hakugyokurou Scarlet_Devil_Mansion 90 Hakurei_Reimu Hakurei_Shrine 0.4 0.1 99 99 99 99 99 99 99 99 Kirisame_Marisa Forest_of_Magic(Marisa's_house) 0.6 0.2 88 77 66 55 44 33 22 11 Izayoi_Sakuya Scarlet_Devil_Mansion 0.4 0.3 77 10 10 10 10 10 10 10 Alice_Margatroid Forest_of_Magic(Alice's_house) 0.4 0.2 66 11 22 33 44 55 66 77 Patchouli_Knowledge Scarlet_Devil_Mansion 0.5 0.3 55 15 15 15 15 15 15 15 Konpaku_Youmu Hakugyokurou 0.6 0.1 44 11 13 17 19 23 29 31 Remilia_Scarlet Scarlet_Devil_Mansion 0.6 0.3 33 99 99 88 88 88 11 11 Saigyouji_Yuyuko Hakugyokurou 0.4 0.1 22 31 29 23 19 17 13 11 Hong_Meiling Scarlet_Devil_Mansion 0.5 0.2 11 22 33 44 55 66 77 88 49
Sample Output
9.06666768697071727374757677787980 484.202020202020202020202020202020
External Links
Author: WU, Zejun
Source: ZOJ Monthly, November 2009
二分套二分。精度问题，样例都没过交一发竟然AC了？！！
#include<cstdio>
#include<map>
#include<queue>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<list>
#include<set>
#include<cmath>
using namespace std;
const int maxn = 1e2 + 5;
const int INF = 1e9;
const double eps = 1e-7;
typedef unsigned long long ULL;
typedef long long LL;
typedef pair<int, double> P;
#define fi first
#define se second
map<string, int> M;
vector<P> G[maxn];
int pos[20];
double V[20], A[20], T[maxn][20];
int sink, q, n;
double C;
double dist[maxn];
struct Node{
int pos;
double dis;
bool operator < (const Node& e) const{
return dis > e.dis;
}
};
double get(int source, double ci, int from){
double v = A[source]*sqrt(ci) + V[source];
priority_queue<Node> q;
while(!q.empty())
q.pop();
for(int i = 0;i < n;i++)
dist[i] = 1e12;
dist[from] = 0;
q.push((Node){from, 0});
while(!q.empty()){
Node tem = q.top();
q.pop();
int pos = tem.pos;
double dis = tem.dis;
if(dis > dist[pos])
continue;
if(pos == sink){
return dis;
}
for(int i = 0;i < G[pos].size();i++){
P& edge = G[pos][i];
int to = edge.fi;
double der = edge.se;
if(dist[to] > dis+der/v+T[to][source]){
dist[to] = dis+der/v+T[to][source];
q.push((Node){to, dist[to]});
}
}
}
}
bool can(double limit){
double left = C;
for(int i = 0;i < q;i++){
double l = 0, r = 1e7, ans = 1e6;
for(int c = 0;c < 100;c++){
double mid = (l+r)/2;
if(limit-get(i, mid, pos[i])>eps){
ans = mid;
r = mid;
}
else
l = mid;
}
left -= ans;
}
if(left < eps)
return false;
return true;
}
int main(){
int m;
while(scanf("%d%d%d", &n, &m, &q) != EOF){
M.clear();
for(int i = 0;i < n;i++)
G[i].clear();
for(int i = 0;i < n;i++){
string s;
cin >> s;
M[s] = i;
if(s == "Hakurei_Shrine")
sink = i;
}
while(m--){
string s;
cin >> s;
int x = M[s];
cin >> s;
int y = M[s];
double dis;
cin >> dis;
G[x].push_back(P(y, dis));
G[y].push_back(P(x, dis));
}
for(int i = 0;i < q;i++){
string s;
cin >> s >> s >> V[i] >> A[i];
pos[i] = M[s];
for(int j = 0;j < n;j++)
cin >> T[j][i];
T[sink][i] = 0;
}
cin >> C;
double l = 0, r = 1e12;
double ans;
for(int c = 0;c < 100;c++){
double mid = (l+r)/2;
if(can(mid)){
r = mid;
ans = mid;
}
else
l = mid;
}
printf("%.7lf\n", ans);
}
return 0;
}
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