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ADVENT: /ad?vent/, n.
The prototypical computer adventure game, first designed by Will Crowther on the PDP-10 in the mid-1970s as an attempt at computer-refereed fantasy gaming, and expanded into a puzzle-oriented game by Don Woods at Stanford in 1976. (Woods had been one of the authors of INTERCAL.) Now better known as Adventure or Colossal Cave Adventure, but the TOPS-10 operating system permitted only six-letter filenames in uppercase. See also vadding, Zork, and Infocom.
It has recently been discovered how to run open-source software on the Y-Crate gaming device. A number of enterprising designers have developed Advent-style games for deployment on the Y-Crate. Your job is to test a number of these designs to see which are winnable.
Each game consists of a set of up to 100 rooms. One of the rooms is the start and one of the rooms is the finish. Each room has an energy value between -100 and +100. One-way doorways interconnect pairs of rooms.
The player begins in the start room with 100 energy points. She may pass through any doorway that connects the room she is in to another room, thus entering the other room. The energy value of this room is added to the player‘s energy. This process continues until she wins by entering the finish room or dies by running out of energy (or quits in frustration). During her adventure the player may enter the same room several times, receiving its energy each time.
The input consists of several test cases. Each test case begins with n, the number of rooms. The rooms are numbered from 1 (the start room) to n (the finish room). Input for the n rooms follows. The input for each room consists of one or more lines containing:
The start and finish rooms will always have enery level 0. A line containing -1 follows the last test case.
In one line for each case, output "winnable" if it is possible for the player to win, otherwise output "hopeless".
5 0 1 2 -60 1 3 -60 1 4 20 1 5 0 0 5 0 1 2 20 1 3 -60 1 4 -60 1 5 0 0 5 0 1 2 21 1 3 -60 1 4 -60 1 5 0 0 5 0 1 2 20 2 1 3 -60 1 4 -60 1 5 0 0 -1
hopeless hopeless winnable winnable
#include <iostream> #include <stack> #include <cstring> #include <cstdio> #include <string> #include <algorithm> #include <queue> #include <set> #include <map> #include <fstream> #include <stack> #include <list> #include <sstream> #include <cmath> using namespace std; /* 老在纠结环怎么处理 参考别人思路：先dfs求出到n的可达点，在spfa求最长路，同时判断正环 */ #define ms(arr, val) memset(arr, val, sizeof(arr)) #define mc(dest, src) memcpy(dest, src, sizeof(src)) #define N 10000 #define M 105 #define INF 0x3fffffff #define vint vector<int> #define setint set<int> #define mint map<int, int> #define lint list<int> #define sch stack<char> #define qch queue<char> #define sint stack<int> #define qint queue<int> qint qi; bool graph[M][M]; bool vis[M];//访问标志 bool reach[M];//reach[i]表示i到n是否可达 int energy[M];//能量值 int cnt[M];//记录入队次数 int dis[M];//最长距离 int n; void dfs(int e)//得到哪些点可达n(逆向搜索) { for (int s = 1; s <= n; s++) { if (graph[s][e] && !vis[s]) { reach[s] = vis[s] = true; dfs(s); } } } bool spfa()//最长路 { bool ans = false; ms(vis, 0); dis[1] = 100; vis[1] = true; qi.push(1); cnt[1]++; int cur; while (!qi.empty()) { cur = qi.front(); qi.pop(); vis[cur] = false; for (int i = 1; i <= n; i++) { if (graph[cur][i] && reach[i] && dis[i] < dis[cur] + energy[i]) {//有边，可达且满足松弛条件 dis[i] = dis[cur] + energy[i]; if (!vis[i]) { cnt[i]++; if (cnt[i] >= n) { ans = true; break; } vis[i] = true; qi.push(i); } } } } while (!qi.empty()) { qi.pop(); } return ans || dis[n] > 0; } void init() { ms(cnt, 0); ms(vis, 0); ms(graph, 0); ms(dis, 0); ms(reach, 0); } int main() { int m, t; while (scanf("%d", &n), n != -1) { init(); for (int i = 1; i <= n; i++) { scanf("%d", &energy[i]); scanf("%d", &m); while (m--) { scanf("%d", &t); graph[i][t] = true; } } reach[n] = 1; dfs(n); if (reach[1] && spfa()) puts("winnable"); else puts("hopeless"); } return 0; }
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标签：des style blog http color os io for
原文地址：http://www.cnblogs.com/jecyhw/p/3914674.html
