Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem D: XYZZY

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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.

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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 energy value for room i
• the number of doorways leaving room i
• a list of the rooms that are reachable by the doorways leaving room i

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".

Sample Input

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

Output for Sample Input

hopeless
hopeless
winnable
winnable

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找到一條 1 到 n 且中間路徑不可小於等於 0, 跟原本負環檢查一樣,  更新次數多於 n 時找到正環,
但此題條件有些不同, 因此找到正環後, 把權值換到無窮大, 之後進行更新, 倘若正環上的點可以連到終點,
則直接輸出 winnable

#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
int map[101][101], mt[101];
int cap[101], arrive[101][101];
void bfs(int n, int st) {
    if(arrive[st][n] == 1)
        return;
    arrive[st][n] = 1;
    int i;
    for(i = 0; i < mt[n]; i++)
        bfs(map[n][i], st);
}
void spfa(int n) {
    if(arrive[1][n] == 0) {
        puts("hopeless");
        return;
    }
    int tn, i, j;
    int dis[101] = {}, used[101] = {};
    int update[101] = {};
    queue<int> Q;
    memset(dis, -1, sizeof(dis));
    dis[1] = 100;
    Q.push(1);
    while(!Q.empty()) {
        tn = Q.front();
        Q.pop();
        used[tn] = 0;
        update[tn]++;
        if(dis[tn] >= 0xfffffff)
            continue;
        if(update[tn] > n) {
            if(arrive[tn][n]) {
                puts("winnable");
                return;
            }
            dis[tn] = 0xfffffff;
        }
        for(i = 0; i < mt[tn]; i++) {
            if(dis[map[tn][i]] < dis[tn] + cap[map[tn][i]]) {
                dis[map[tn][i]] = dis[tn] + cap[map[tn][i]];
                if(used[map[tn][i]] == 0) {
                    used[map[tn][i]] = 1;
                    Q.push(map[tn][i]);
                }
            }
        }
    }
    if(dis[n] > 0)
        puts("winnable");
    else
        puts("hopeless");
}
int main() {
    int n, m, i, y;
    while(scanf("%d", &n) && n > 0) {
        memset(mt, 0, sizeof(mt));
        memset(arrive, 0, sizeof(arrive));
        for(i = 1; i <= n; i++) {
            scanf("%d %d", &cap[i], &m);
            while(m--) {
                scanf("%d", &y);
                map[i][mt[i]++] = y;
            }
        }
        for(i = 1; i <= n; i++)
            bfs(i, i);
        spfa(n);
    }
    return 0;
}