POJ1042 Gone Fishing

采用贪心策略。

假设他从1湖泊走到x 湖泊，这还剩下 h*12 - sigma(T1--Tx-1)。(单位时间为5分钟)。然后再用剩下的时间去钓1-x的湖泊的鱼。 每次都选择最多鱼的湖泊钓。

code：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn = 30;
 
int f[maxn], tf[maxn], d[maxn], t[maxn], path[maxn];
int ans, p[maxn];
 
int main()
{
    int  h, n, i, j, k;
    while(scanf("%d",&n),n) {
        scanf("%d",&h);
        h *= 12;
        for(i=1; i<=n; ++i) scanf("%d",&f[i]);
        for(i=1; i<=n; ++i) scanf("%d",&d[i]);
        for(i=1; i<=n-1; i++) scanf("%d",&t[i+1]);
        for(i=2,t[1]=0; i<=n; ++i) t[i] += t[i-1];
        ans = -1000;
        for(k=1; k<=n; ++k)
            if(h>t[k]) {
                int th = h - t[k];
                int sum = 0;
                memset(path, 0, sizeof path );
                for(i=1; i<=n; ++i) tf[i] = f[i];
                while(th>0) {
                    int maxx  = 0;
                    int mark = 0;
                    for(i=1; i<=k; ++i)
                        if(maxx<tf[i]) {
                            maxx = tf[i];
                            mark  = i;
                        }
                    if(!mark) break;
                    sum += maxx;
                    tf[mark] -= d[mark];
                    path[mark]++;
                    th--;
                }
                path[1] += th;
                if(ans<sum) {
                    ans = sum;
                    for(j=1; j<=n; j++)
                        p[j] = path[j];
                }
            }
        for(i=1; i<n; ++i) printf("%d, ",p[i]*5);
        printf("%d\n",p[i]*5);
        printf("Number of fish expected: %d\n\n", ans);
    }
    return 0;
}