uva10382

来源：互联网发布：python 实现卷积编辑：程序博客网时间：2024/06/06 20:57

题目大意：
最少要用多少个给定的圆才能把长度为l，宽度为w的完全覆盖住。

思路：
区间覆盖问题，本来一直WA和TLE，结果现在寒假训练找又做到了，改了一个地方，居然A了。

代码：

#include <iostream>using namespace std;#include <cstring>#include <stdio.h>#include <cmath>int N;double l,w;struct node {    double pos;    double r;}n[10005];int main() {    int count1,index;    double _max;    double p,r;    while(scanf("%d%lf%lf",&N,&l,&w)!= EOF) {        index = 0;        count1 = 0;        _max = 0;        for(int i = 0; i < N; i++) {            scanf("%lf %lf",&p,&r);            if(w/2 >= r)                continue;            double t = sqrt(r*r-w*w/4);            n[index].pos = p-t;            n[index].r = p+t;            index++;        }        double min = 0,max = 0;        int rr = 0,pp,t;        while(1) {            if(min >= l)                break;        //  max = 0;            rr =0;            for(int i = 0 ; i < index; i++) {                if(n[i].pos  <= min && n[i].r > max ) {                    max =  n[i].r;                    rr = 1;                    pp = i;                }            }            if(rr) {                min = n[pp].r;                count1++;            }            else                 break;            }        if(rr)             printf("%d\n",count1);        else            printf("%d\n",-1);    }    return 0;   }

#include <iostream>using namespace std;#include <cstring>#include <stdio.h>#include <cmath>int N;double l,w;struct node {    double pos;    double r;}n[10005];bool judge(double a,double b,double c) {//  printf("%d %d %d \n",a,b,c);//  printf("%lf",(sqrt((a+b+c)*(a+b-c)*(a+c-b)*(b+c-a))/c));    if(sqrt((a+b+c)*(a+b-c)*(a+c-b)*(b+c-a))/c >=w)        return true;    return false;}int main() {    int count1;    double _max;    while(scanf("%d%lf%lf",&N,&l,&w)!= EOF) {        count1 = 0;        _max = 0;        for(int i = 0; i < N; i++) {            scanf("%lf %lf",&n[i].pos,&n[i].r);            if(n[i].pos + n[i].r > _max)                _max = n[i].pos + n[i].r;        }        if(_max < l) {            printf("%d\n",-1);        }        else {            double min = 0,max = 0;            int rr = 0,p,t;            bool first = true;        while(1) {    //      cout << 1;            if(min >= l)                break;            max = 0;            rr =0;            for(int i = 0 ; i < N; i++) {                if(n[i].pos - n[i].r <= min && n[i].pos + n[i].r > max ) {                if(first){                //  first = false;                    max = n[i].pos + n[i].r;                    rr = 1;                    p = i;                }                else if(judge(n[t].r,n[i].r,n[i].pos-n[t].pos) ) {                        max = n[i].pos + n[i].r;                        rr = 1;                        p = i;                    }                }            }            if(rr) {                first = false;                min = n[p].pos + n[p].r;                t = p;                count1++;            }            else                 break;            }        if(rr)             printf("%d\n",count1);        else            printf("%d\n",-1);        }    }    return 0;   }

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