uva10557

题意:

初始值为100,给出每个房间的能量值,求能否在能量值为正的情况下从1走到n

思路:

floyd和dfs,这道题用了两种剪枝,一个是到当前房间能量值控制在最优,否则不再进入这个房间,一个是用floyd判断当前点是否可达终点

还有一个重要的剪枝就是出现正环则为可以

此题还可以用bfs做,由于BFS不会出现死循环,所以实际上是可以不考虑回环问题的,但是bfs比dfs要耗时的多,用到队列存储,直到到达最后一个地方为正才结束,否则就一直到无路可走,这显然耗时的多

dfs:

#include<cstdio>#include<cstring>#include<iostream>using namespace std;int map[105][105];int reach[105][105];int en[105];int n;int flag;int vis[105];int env[105];void floyd() {for(int k=1; k<=n; k++)for(int i=1; i<=n; i++)for(int j=1; j<=n; j++) {if(reach[i][k] && reach[k][j]) reach[i][j] = 1;}reach[n][n] = 1;}void dfs(int cur, int sum) {if(flag) return;if(cur==n) {if(sum >= 0) {flag = 1;return;}}for(int i=1; i<=n; i++) {if(map[cur][i] && sum+en[i]>0 && reach[i][n] && env[i]<sum+en[i]) {vis[i]++;if(vis[i]>1 && en[i]>0) {flag = 1;return;}env[i] = sum+en[i];dfs(i, sum+en[i]);vis[i]--;}}}int main()  {int temp;int m;while(scanf("%d", &n) && n != -1) {memset(vis, 0, sizeof(vis));memset(map, 0, sizeof(map));memset(reach, 0, sizeof(reach));memset(env, -1000000, sizeof(env));for(int i=1; i<=n; i++) {scanf("%d %d", &en[i], &m);for(int j=0; j<m; j++) {scanf("%d", &temp);map[i][temp] = 1;reach[i][temp] = 1;}}floyd();flag = 0;vis[1] = 1;if(reach[1][n])dfs(1, 100);if(flag) {printf("winnable\n");}else printf("hopeless\n");}return 0;}

bfs:

#include<cstdio>#include<cstring>#include<iostream>#include<queue>using namespace std;int map[105][105];int reach[105][105];int en[105];int n;int flag;int vis[105];int env[105];void floyd() {for(int k=1; k<=n; k++)for(int i=1; i<=n; i++)for(int j=1; j<=n; j++) {if(reach[i][k] && reach[k][j]) reach[i][j] = 1;}reach[n][n] = 1;}void bfs() {queue<int>q;q.push(1);while(!q.empty()) {int cur = q.front();q.pop();for(int i=1; i<=n; i++) {if(reach[i][n] && map[cur][i] && env[i]<env[cur]+en[i] && env[cur]+en[i]>0) {env[i] = env[cur]+en[i];q.push(i);if(env[n]>0) {flag = 1;return;}}}}}int main()  {int temp;int m;while(scanf("%d", &n) && n != -1) {memset(vis, 0, sizeof(vis));memset(map, 0, sizeof(map));memset(reach, 0, sizeof(reach));memset(env, -1000000, sizeof(env));for(int i=1; i<=n; i++) {scanf("%d %d", &en[i], &m);for(int j=0; j<m; j++) {scanf("%d", &temp);map[i][temp] = 1;reach[i][temp] = 1;}}floyd();flag = 0;vis[1] = 1;env[1] = 100;if(reach[1][n])bfs();if(flag) {printf("winnable\n");}else printf("hopeless\n");}return 0;}