UVA - 10557 XYZZY（DFS + BFS）
XYZZY
| Time Limit:1000MS | Memory Limit:32768KB | 64bit IO Format:%I64d & %I64u |
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Description
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 deploymenton 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 ofthe rooms is the finish. Each room has an energy value between -100 and +100. One-waydoorways interconnect pairs of rooms.
The player begins in the start room with 100 energy points. She may pass through anydoorway 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 untilshe 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 energyeach time.
Input
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 nrooms follows. The input for each room consists of one or more lines containing:
1. the energy value for room i;
2. the number of doorways leaving room i;
3. a list of the rooms that are reachable by the doorways leaving room i.
The start and finish rooms will always have energy level 0. A line containing -1 followsthe last test case.
Output
In one line for each case, output "winnable" if it is possible for the player to win, otherwiseoutput "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
Sample Output
hopeless hopeless winnable winnable
题目大意：
给你的第一行的数字是节点数。 节点数不超过100
然后要求你输入每个节点的能量值，和出度，以及出度所对应的节点。
然后初始化的能量值为100，你每走一个节点都会加上该节点对应的能量值，如果当前的能量值小于等于0,就game over
解析：
这题做了好久，其间忘记了初始化WA了好多次，看了网络上的题解才水过的。
主要是先是用DFS，判断是否存在一条每步的能量值大于0的路线，如果存在继续DFS。
还要考虑是否存在正环，如果存在正环的话能无限刷能量，此时用BFS，判断存在正环的点是否能到达终点，如果能到达终点就win。
#include <stdio.h>
#include <queue>
#include <string.h>
using namespace std;
const int N = 105;
int E[N];
int v[N];
int edge[N][N];
int n;
bool bfs(int pos) {
bool vis[N] = {0};
queue<int> q;
q.push(pos);
vis[pos] = true;
while( !q.empty()) {
int front = q.front();
q.pop();
for(int i = 1; i <= n; i++) {
if( !vis[i] && edge[front][i]) {
if( i == n) {
return true;
}
vis[i] = true;
q.push(i);
}
}
}
return false;
}
bool dfs(int pos,int val) {
if(pos == n) {
return true;
}
for(int i = 1; i <= n; i++) {
// 如果该点已经访问过，且满足这次访问该点的能量，大于上次访问该点的能量
// 则表明构成正环，即可以无限获得能量，直接搜索该点能不能到达终点
if(edge[pos][i] && v[i] + val > 0) {
if( !E[i] ) {
E[i] = val + v[i];
if( dfs(i,E[i]) ) {
return true;
}
} else if( v[i] + val > E[i] && bfs(i)) {
return true;
}
}
}
return false;
}
int main() {
while(scanf("%d",&n) != EOF && n+1) {
memset(edge,0,sizeof(edge));
int m,t;
for(int i = 1; i <= n; i++) {
scanf("%d%d",&v[i],&m);
for(int j = 1; j <= m; j++) {
scanf("%d",&t);
edge[i][t] = 1;
}
}
memset(E,0,sizeof(E));
if(dfs(1,100)) {
printf("winnable\n");
}
else {
printf("hopeless\n");
}
}
return 0;
}