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include<cstdio> include<iostream> include<cstdlib> include<cstring> include<queue> usingnamespacestd; structnode { intx,y,step; };//记录点的坐标以及到该点的时间 intn,m; charmap[209][209]; intdx[4]={0,-1,0,1}; intdy[4]={-1,0,1,0}; intvis[209][209],num[209][209];//vis标...
include<cstdio>//一开始找规律，虽然总感觉能找到但是很浪费时间其实，然后看到了BFS简直豁然开朗 include<cstring> include<algorithm> include<queue> defineN100+5 usingnamespacestd; structnode{ inta,b,s,t;//a是大杯，b是小杯，s是可乐，t代表次数 }st;//深刻理解了BFS不止是迷宫图什么的，只要和最短最少有关就可以考虑，只要有起点和终点，有搜索“方向”就可以搜索 inta,b,s; intvis[N][N]; i...
include<stdio.h>//也是BFS的模板题，三个搜索方向，分别判断，挺好想到的 include<string.h> include<queue> usingnamespacestd; constintN=1000000; intmap[N+10]; intn,k; structnode { intx,step; }; intcheck(intx) { if(x<0||x>=N||map[x])//这个判断的顺序很重要，先得判断有没有越界，不然runtimeerror return0; return1; } int...
include<cstdio> include<stack> include<algorithm> include<queue> include<cstring> usingnamespacestd; intmaze[10][10]; intd[10][10]; intvisit[10][10]; intdx[4]={-1,1,0,0};//表示上下左右四个方向 intdy[4]={0,0,-1,1}; structNode{ intx,y; Node(intr,intc):x(r),y(c){}; Node(){} }p...
include<iostream> include<algorithm> usingnamespacestd; intweight[400];//简单的贪心，只要排序后体重小的去和体重大的一起就好了 intmain() { intN; intn,w; while(cin>>N) { while(N--) { while(cin>>w>>n) { intcount=0; for(inti=0;i<n;i) cin>>weight[i]; sort(w...
思路： 贪心思想（一般都是先排序） 关键步骤：每次从此岸到彼岸移动的两个人要么这两个人中有一个是最快的那个人，要么这两个人到达彼岸后再也不回来。即：要么最快+最慢，要么最慢+次慢。 1.对N个人过河时间从小到大排序。speed[i] 2.分情况讨论： ⑴当n=1，直接过河。sum=speed[0] (2)当n=2，直接过河。sum=speed[1] (3)当n=3，无论怎么过河，sum=speed[0]+speed[1]+speed[2] (4)当n=4，设从小到大排序后位a,b,c,d 用最小的来送：b+a+c+a+d=2a+b+c+d（a,b过去，a回来，a,...
include<iostream>//经典贪心会议安排 include<algorithm> include<cmath> usingnamespacestd; structF{ doublebegin; doubleend; }s[10001]; boolcmp(Fa,Fb) { returna.end<b.end;//按照结束时间排序 } intmain() { intT; intN; while(cin>>T) { while(T--) { while(cin>>N) { ...
include<iostream>//dp问题，求一个区间覆盖问题的贪心，和会议安排是不同的，其实核心和最长公共子序列比较像 include<algorithm> include<cmath> usingnamespacestd; structF{ doublex; doubler; doublel; doublebegin; doubleend; }s[10000]; boolcmp1(Fa,Fb) { returna.r>b.r; } boolcmp2(Fa,Fb) { returna.begin<b.begin; } ...
include<iostream> include<algorithm> include<cmath>//贪心问题，圆形面积转化分析，实际有效面积其实是一个矩形，所以用矩形的长度填满了目标长度即可 usingnamespacestd; structF{ doubler; doublel; }s[10000]; boolcmp(Fa,Fb) { returna.l>b.l; } intmain() { intT; intN; while(cin>>T) { while(T--) { while(cin>...
include<iostream> include<algorithm> usingnamespacestd; structwood{ intlen; intwei; boolv;//用来标记防止重复访问 }w[5001]; boolcmp(wooda,woodb) { if(a.lenb.len) returna.wei>b.wei;//len是第一标准 else returna.len>b.len; } intmain() { intT,N,sum; while(cin>>T) { while(T--) ...
include<cstdio> include<cstring> include<algorithm> usingnamespacestd;//这是区间最大重载数问题 intmain(){ intt,n,count[410],i,start,end,k; scanf("%d",&t); while(t--){ scanf("%d",&n); memset(count,0,sizeof(count)); while(n--){ scanf("%d%d",&start,&end); //可能出发...
include<iostream> include<algorithm> include<iomanip> usingnamespacestd; charmap[5][5]; intres; intn; booljudge(intx,inty) { if(map[x][y]!='.') returnfalse; for(inti=x1;i>=0;i--) { if(map[i][y]'X') break; if(map[i][y]'O') returnfalse; } for(inti=y1;i>=0;i-...
include"stdafx.h" include<graphics.h> include<conio.h> include<stdio.h> include<stdlib.h> //引用WindowsMultimediaAPI pragmacomment(lib,"Winmm.lib") define Width800 defineHigh600 intmap_num=1; intHero_x,Hero_y;//主角坐标 intboard_x[13],board_y[13];//平台坐标 intsuper_board_x,supe...
include<conio.h> include<graphics.h> include<windows.h> defineHigh480//游戏画面尺寸 defineWidth640 //全局变量 intball_x,ball_y;//小球的坐标 intball_vx,ball_vy;//小球的速度 intradius;//小球的半径 intbar1_left,bar1_right,bar1_top,bar1_bottom;//挡板1的上下左右位置坐标 intbar2_left,bar2_right,bar2_top,bar2_bottom;//挡板...
include<graphics.h> include<conio.h> include<math.h> include<stdio.h> //引用WindowsMultimediaAPI pragmacomment(lib,"Winmm.lib") defineHigh700//游戏画面尺寸 defineWidth580 IMAGEimg_bk;//背景图片 floatposition_x,position_y;//飞机位置 floatbullet_x,bullet_y;//子弹位置 floatenemy_x,enemy_y;//敌...
include<graphics.h> include<conio.h> include<stdlib.h> include<stdio.h> pragmacomment(lib,"Winmm.lib") IMAGEimg_bk,img_bd1,img_bd2,img_bar_up1,img_bar_up2,img_bar_down1,img_bar_down2; intbird_x; intbird_y; intbar_x; intbar_up_y,bar_down_y; booldead; intscore; intv=1; vo...