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;

}