UVa 757 – Gone Fishing

【題目連結】

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=113&page=show_problem&problem=698

【原題】

John is going on a fishing trip. He has h hours available ( ),
and there are n lakes in the area ( ) all reachable along a single, one-way road. John starts at lake 1, but he
can finish at any lake he wants. He can only travel from one lake to the next one, but he does not have to stop at any lake unless he wishes to. For each ,
the number of 5-minute intervals it takes to travel from lake i to lake i + 1 is denoted ti ( ).
For example, t3 = 4 means that it takes 20 minutes to travel from lake 3 to lake 4.

To help plan his fishing trip, John has gathered some information about the lakes. For each lake i, the number of fish expected to be caught in the initial 5 minutes, denoted fi ( ),
is known. Each 5 minutes of fishing decreases the number of fish expected to be caught in the next 5-minute interval by a constant rate of di ( ).
If the number of fish expected to be caught in an interval is less than or equal to di, there will be no more fish left in the lake in the next interval. To simplify the planning, John assumes that no one else will be fishing at
the lakes to affect the number of fish he expects to catch.

Write a program to help John plan his fishing trip to maximize the number of fish expected to be caught. The number of minutes spent at each lake must be a multiple of 5.

Input 

You will be given a number of cases in the input. Each case starts with a line containing n.
This is followed by a line containing h. Next, there is a line of n integers
specifying f_{i ( ),
then a line of n integers di ( ),
and finally, a line of n - 1 integers ti ( ).
Input is terminated by a case in which n = 0.}

Output 

For each test case, print the number of minutes spent at each lake, separated by commas, for the plan achieving the maximum number of fish expected to be caught (you should print the entire plan on one
line even if it exceeds 80 characters). This is followed by a line containing the number of fish expected. If multiple plans exist, choose the one that spends as long as possible at lake 1, even if no fish are expected to be caught in some intervals. If there
is still a tie, choose the one that spends as long as possible at lake 2, and so on. Insert a blank line between cases.

Sample Input 

2110 12 524410 15 20 170 3 4 31 2 34410 15 50 300 3 4 31 2 30

Sample Output 

45, 5Number of fish expected: 31240, 0, 0, 0Number of fish expected: 480115, 10, 50, 35Number of fish expected: 724

【分析與總結】

超級經典的一道貪心題，LRJ黑書上貪心章節的第一道例題。

我是用堆來實現的。

【代碼】

/* * UVa: 757 - Gone Fishing * Greedy * Time: 0.044s(UVa), 79MS(poj) * Author: D_Double * */#include<iostream>#include<cstring>#include<cstdio>#include<queue>#define MAXN 30using namespace std;int h,n;int ans[MAXN], tmp[MAXN];struct Node{    int no;        // 第幾號湖    int rate;      // 每5分鐘釣的魚    int down;      // 每5分鐘減少的釣魚數量    int time;      // 從第一個湖走到當前這個湖的時間    friend bool operator < (const Node&a,const Node&b){        if(a.rate!=b.rate)            return a.rate<b.rate;        return a.no>b.no; // 如果每5分鐘釣的魚相同，優先序號釣小的湖    }}arr[MAXN];priority_queue<Node>que;void greedy(){    int maxSum=-10000;    for(int i=0; i<n; ++i){ // 枚舉釣1~i個湖的情況        while(!que.empty()) que.pop();        for(int j=0; j<=i; ++j) que.push(arr[j]);        int leftTime=h*60-arr[i].time, sum=0;        memset(tmp, 0, sizeof(tmp));        while(leftTime > 0){            Node temp=que.top();            que.pop();            if(temp.rate<=0) break;            sum += temp.rate;            temp.rate -= temp.down;            tmp[temp.no] += 5;            que.push(temp);            leftTime -= 5;        }          if(leftTime>0) tmp[0] += leftTime; // 注意把剩下的時間都加到第一個湖上        if(sum > maxSum){            maxSum=sum;            for(int j=0; j<n; ++j)                ans[j]=tmp[j];        }    }    printf("%d",ans[0]);    for(int i=1; i<n; ++i)        printf(", %d",ans[i]);    printf("\n");    printf("Number of fish expected: %d\n", maxSum);}int main(){    bool flag=false;    while(~scanf("%d",&n)&&n){        scanf("%d",&h);        for(int i=0; i<n; ++i){            scanf("%d",&arr[i].rate);            arr[i].no = i;        }        for(int i=0; i<n; ++i){            scanf("%d",&arr[i].down);        }        arr[0].time=0;        for(int i=1; i<=n-1; ++i){            int t;            scanf("%d",&t);            arr[i].time = arr[i-1].time+t*5;        }        if(flag)printf("\n");        else flag=true;        greedy();     }    return 0;}

——  生命的意義，在於賦予它意義。
               原創 http://blog.csdn.net/shuangde800 ， By   D_Double  (轉載請標明)