100000584 - 《算法笔记》4.4小节——算法初步->贪心

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《算法笔记》4.4小节——算法初步->贪心
用局部最优推断全局最优
贪心算法使用的问题一定满足最优子结构性质，即一个问题的最优解可以由其子问题的最优解有效构造
并不是所有问题都适用贪心算法

1、简单贪心
PTA B1020 月饼 (25分)

#include <cstdio>
#include <algorithm>
using namespace std;

struct mooncake{
    double store;
    double sell;
    double price;
}cake[1010];

bool cmp(mooncake a,mooncake b){
    return a.price> b.price;
}

int main(){
    int n;
    double D;
    scanf("%d%lf",&n,&D);
    for(int i=0;i<n;i++){
        scanf("%lf",&cake[i].store);
    }
    for(int i=0;i<n;i++){
        scanf("%lf",&cake[i].sell);
        cake[i].price=cake[i].sell/cake[i].store;
    }
    
    sort(cake,cake+n,cmp);
    double ans=0;
    for(int i=0;i<n;i++){
        if(cake[i].store<=D){
            D-=cake[i].store;
            ans+=cake[i].sell;
        }else{
            ans+=cake[i].price*D;
            break;
        }
    }
    printf("%.2f",ans);
    return 0;
}

PTA B1023 组个最小数 (20分)

#include <cstdio>

int main(){
    int count[10];
    for(int i=0;i<10;i++){
        scanf("%d",&count[i]);
    }
    for(int i=1;i<10;i++){
        if(count[i]>0){
            printf("%d",i);
            count[i]--;
            break;
        }
    }
    for(int i=0;i<10;i++){
        for(int j=0;j<count[i];j++){
            printf("%d",i);
        }
    }
    return 0;
}

2、区间贪心
区间不相交问题：a选择被包含的区间，b按左端点从大到小排列则删去区间包含情况总是选择左端点最大或右端点最小
区间选点问题：多个闭区间中最少需要确定多少个点才能是每个闭区间都至少存在一个点——a选择被包含的区间，b按照左端点从大到小排列则删去区间包含情况总是选择左端点最大的区间的左端点

#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn = 110;

struct Inteval {
	int x, y;
}I[maxn];

bool cmp(Inteval a, Inteval b) {
	if (a.x != b.x) return a.x > b.x;	//左端点从大到小
	else return a.y > b.y;	//左相同 右端点从小到大
}

int main() {
	int n;
	while (scanf("%d", &n), n != 0) {
		for (int i = 0; i < n; i++) {
			scanf("%d%d", &I[i].x, &I[i].y);
		}

		sort(I, I + n, cmp);
		//lastX=I[i].x即解决区间选点问题
		int ans = 1, lastX = I[0].x;	//	lastX记录上一个被选中的区间的左端点
		for (int i = 0; i < n; i++) {
			if (I[i].y <= lastX) {
				lastX = I[i].x;
				ans++;
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}

1126 Problem A 看电视

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

struct program {
	int si;
	int ei;
}pgm[110];

bool cmp(program a,program b) {
	if (a.si == b.si) return a.ei < b.ei;
	else  return a.si > b.si;
}

int main() {
	int n;
	while (scanf("%d", &n) != EOF && n != 0) {
		for (int i = 0; i < n; i++) {
			scanf("%d%d", &pgm[i].si,&pgm[i].ei);
		}
		sort(pgm, pgm + n, cmp);
		int lasttime = pgm[0].si;
		int cnt = 1;
		for (int i = 1; i < n; i++) {
			if (pgm[i].ei <= lasttime) {
				cnt++;
				lasttime = pgm[i].si;
			}
		}
		printf("%d\n", cnt);
	}
	return 0;
}

1128 Problem B 出租车费

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int main() {
	int n;
	while (scanf("%d", &n) != EOF && n != 0) {
		double fee = 0;
		//写好分段函数 也可再直观画图
		if (n <= 4)
			fee = 10;
		else if (n > 4 && n <= 8)
			fee = 10 + (n - 4) * 2;
		else {
			if ((n % 8) <= 4) fee = (n / 8) * 18 + (n % 8)*2.4;
			else fee = (n / 8) * 18 + (10 + (n % 8 - 4) * 2);
		}
		//注意输出格式，用fee-(int)fee==0或fee==(int)fee判断是否是整数
		if(fee-(int)fee==0) printf("%d\n", (int)fee);
		else printf("%.1f\n", fee);
	}

	return 0;
}

2031 Problem C To Fill or Not to Fill
PAT 1033 To Fill or Not to Fill (25分)
有两个测试点未通过

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

struct gasstation {
	double pi;
	double di;
}gs[510];

bool cmp(gasstation a, gasstation b) {
	if (a.di != b.di)	return a.di < b.di;
	else return a.pi < b.pi;
}

int main() {
	double cmax, D, davg;
	int N;
	while (scanf("%lf %lf %lf %d", &cmax, &D, &davg, &N) != EOF) {
		for (int i = 0; i < N; i++) {
			scanf("%lf %lf", &gs[i].pi, &gs[i].di);
		}

		double price = 0;
		double discur = 0, ccur = 0, dismax = cmax * davg;
		int ncur = 0, unreachable = 0;
		sort(gs, gs + N, cmp);
		if (gs[0].di != 0) {
			printf("The maximum travel distance = %.2lf", discur);
			unreachable = 1;
		}
		else {
			while (ncur < N) {
				if (ncur == N - 1) {
					if (((D - gs[ncur].di) / davg) > cmax) {
						printf("The maximum travel distance = %.2lf", discur);
						unreachable = 1;
					}
					else if (ccur >= (D - gs[ncur].di) / davg)
						discur = D;
					else  price += (D - gs[ncur].di) / davg * gs[ncur].pi;
					break;
				}
				if (gs[ncur + 1].di > (discur + dismax)) {
					price += gs[ncur].pi*(cmax - ccur);
					discur = gs[ncur].di + cmax / davg;
					printf("The maximum travel distance = %.2lf\n", discur);
					unreachable = 1;
					break;
				}
				int flag = 0;
				double pmin = gs[ncur].pi;
				int next = ncur + 1;

				for (int i = next; i < N && gs[i].di < (gs[ncur].di + dismax); i++) {
					if (gs[i].pi < pmin) {
						pmin = gs[i].pi;
						next = i;
						flag = 1;
						break;
					}
				}//找后续加油站中第一个比当前便宜的
				if (flag == 0) {
					pmin = gs[ncur + 1].pi;
					for (int i = next; i < N && gs[i].di < (gs[ncur].di + dismax); i++) {
						if (gs[i].pi <= pmin) {
							pmin = gs[i].pi;
							next = i;
						}
					}
				}
				//如果没有就找后续中最便宜的
				if (gs[next].pi > gs[ncur].pi) {
					price += (cmax - ccur)*gs[ncur].pi;
					discur = gs[next].di;
					ccur = cmax - (gs[next].di - gs[ncur].di) / davg;
					ncur = next;
				}//比当前加油站贵，就加满
				else {
					if (ccur > (gs[next].di - gs[ncur].di) / davg) ncur = next;
					else {
						price += ((gs[next].di - gs[ncur].di) / davg - ccur) * gs[ncur].pi;
						discur = gs[next].di;
						ccur = 0;
						ncur = next;
					}
				}//如果比当前加油站便宜，加够跑到下一站
			}
			if (unreachable == 0)
				printf("%.2lf", price);
		}
	}
	return 0;
}

2132 Problem D Repair the Wall

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

bool cmp(int a, int b) {
	return a > b;
}

int main() {
	int l, n;
	while (scanf("%d %d", &l, &n) != EOF) {
		int stone[610];
		for (int i = 0; i < n; i++)
			scanf("%d", &stone[i]);
		sort(stone, stone + n, cmp);

		int cnt = 0, flag = 0;
		for (int i = 0; i < n; i++) {
			if (l <= stone[i]) {
				cnt++;
				flag = 1;
				break;
			}
			else {
				l -= stone[i];
				cnt++;
			}
		}
		if (flag == 1) printf("%d\n", cnt);
		else printf("impossible\n");
	}
	return 0;
}

2134 Problem E FatMouse’s Trade

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

struct mouse {
	double j;
	double f;
	double a;
};

bool cmp(mouse s,mouse q) {
	return s.a > q.a;
}

int main() {
	double m;
	int n;
	while (scanf("%lf %d", &m, &n) != EOF && m != -1 && n != -1) {
		mouse mou[1010];
		for (int i = 0; i < n; i++) {
			scanf("%lf %lf", &mou[i].j, &mou[i].f);
			if (mou[i].f != 0)
				mou[i].a = mou[i].j / mou[i].f;
			else
				mou[i].a = 1000;
		}
		double aqr = 0.0;
		sort(mou, mou + n, cmp);
		for (int i = 0; i < n; i++) {
			if (m == 0)
				break;
			if (mou[i].f <= m) {
				m -= mou[i].f;
				aqr += mou[i].j;
			}
			else {
				aqr += m*1.0 / mou[i].f * mou[i].j;
				m = 0;
			}
		}
		printf("%.3lf\n", aqr);
	}
	return 0;
}

2143 Problem F 迷瘴

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

bool cmp(double a,double b) {
	return a < b;
}

int main() {
	int c;
	scanf("%d", &c);
	while (c--) {
		int n, v;
		double sampw=0, testw=0;
		scanf("%d %d %lf", &n, &v, &sampw);
		double p[110];
		int cnt;
		for (int i = 0; i < n; i++) 
			scanf("%lf", &p[i]);
		sort(p, p + n, cmp);
		for (cnt = 0; cnt < n; cnt++) {
			double temp = 1.0*(testw*cnt + p[cnt]) / (cnt + 1);
			if (temp <= sampw) testw = temp;
			else break;
		}
		if (cnt == 0)
			printf("0 0.00\n");
		else
			printf("%d %.2lf\n", cnt*v, testw*0.01);
	}
	return 0;
}

5038 Problem G 找零钱

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

struct money {
	int coin;
	int num;
};


bool cmp(money a, money b)
{
	if (a.num == 0 && b.num != 0)
		return false;
	else if (a.num != 0 && b.num == 0)
		return true;
	else if (a.num != 0 && b.num != 0)
		return a.coin > b.coin;
	else
		return true;
}

int main() {
	int n;
	while (scanf("%d", &n) != EOF) {
		money mny[5] = { {50,0},{20,0},{10,0},{5,0},{1,0} };
		for (int i = 0; i < 5; i++) {
			if (n >= mny[i].coin) {
				mny[i].num = n / mny[i].coin;
				n = n % mny[i].coin;
			}
		}
		sort(mny, mny + 5, cmp);
		for (int i = 0; i < 5; i++) {
			if (mny[i].num != 0) {
				printf("%d*%d", mny[i].coin, mny[i].num);
				if (mny[i + 1].num != 0 && i < 4) printf("+");
			}
			
		}
		printf("\n");
	}
	return 0;
}