4.3.7 Kill the monster

Kill the monster

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 118    Accepted Submission(s): 86

Problem Description There is a mountain near yifenfei’s hometown. On the mountain lived a big monster. As a hero in hometown, yifenfei wants to kill it. Now we know yifenfei have n spells, and the monster have m HP, when HP <= 0 meaning monster be killed. Yifenfei’s spells have different effect if used in different time. now tell you each spells’s effects , expressed (A ,M). A show the spell can cost A HP to monster in the common time. M show that when the monster’s HP <= M, using this spell can get double effect.

Input The input contains multiple test cases. Each test case include, first two integers n, m (2<n<10, 1<m<10^7), express how many spells yifenfei has. Next n line , each line express one spell. (Ai, Mi).(0<Ai,Mi<=m).

Output For each test case output one integer that how many spells yifenfei should use at least. If yifenfei can not kill the monster output -1.

Sample Input 3 100 10 20 45 89 5 40 3 100 10 20 45 90 5 40 3 100 10 20 45 84 5 40

Sample Output 3 2 -1 #include<cstdio> #include<cstring> #include<stack> using namespace std; const int MAXN = 10; int a[MAXN], m[MAXN]; int n, HP; struct node{ int hp, t; bool vis[MAXN]; }; int dfs(){ int mint = 1e6; stack<node> s; node cur, next; memset(cur.vis, 0, sizeof(cur.vis)); cur.hp = HP; cur.t = 0; while(!s.empty()) s.pop(); s.push(cur); while(!s.empty()){ cur = s.top(); s.pop(); for(int i = 0; i < n; i++){ next = cur; if(!next.vis[i]){ if(next.hp <= m[i]){ next.hp -= 2 * a[i]; } else next.hp -= a[i]; next.t++; next.vis[i] = 1; if(next.hp <= 0){ if(mint > next.t) mint = next.t; continue; } s.push(next); } } } if(mint == 1e6) return -1; else return mint; } int main(){ while(scanf("%d%d", &n, &HP) != EOF){ for(int i = 0; i < n; i++){ scanf("%d%d", &a[i], &m[i]); } int ans = dfs(); printf("%d\n", ans); } return 1; }