UVa 757 - Gone Fishing
【题目链接】
【原题】
John is going on a fishing trip. He hashhours available (), and there arenlakes 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 lakeito lakei+ 1 is denotedti(). 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 lakei, the number of fish expected to be caught in the initial 5 minutes, denotedfi(), 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 ofdi(). If the number of fish expected to be caught in an interval is less than or equal todi, 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 d i ( ), and finally, a line of n - 1 integers t i ( ). 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
2 1 10 1 2 5 2 4 4 10 15 20 17 0 3 4 3 1 2 3 4 4 10 15 50 30 0 3 4 3 1 2 3 0
Sample Output
45, 5 Number of fish expected: 31 240, 0, 0, 0 Number of fish expected: 480 115, 10, 50, 35 Number of fish expected: 724
【分析与总结】
超级经典的一道贪心题，LRJ黑书上贪心章节的第一道例题。
我是用堆来实现的。
【代码】
/*
* UVa: 757 - Gone Fishing
* Greedy
* Time: 0.044s(UVa), 79MS(poj)
* Author: D_Double
*
*/
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define MAXN 30
using namespace std;
int h,n;
int ans[MAXN], tmp[MAXN];
struct Node{
int no; // 第几号湖
int rate; // 每5分钟钓的鱼
int down; // 每5分钟减少的钓鱼数量
int time; // 从第一个湖走到当前这个湖的时间
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分钟钓的鱼相同，优先序号钓小的湖
}
}arr[MAXN];
priority_queue<Node>que;
void greedy(){
int maxSum=-10000;
for(int i=0; i<n; ++i){ // 枚举钓1~i个湖的情况
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; // 注意把剩下的时间都加到第一个湖上
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;
}
—— 生命的意义，在于赋予它意义。
原创http://blog.csdn.net/shuangde800，By D_Double (转载请标明)