HDU 3262 Seat taking up is tough (模拟搜索)
传送门：http://acm.hdu.edu.cn/showproblem.php?pid=3262
题意：教室有n*m个座位，每个座位有一个舒适值，有K个学生在不同时间段进来，要占t个座位，必须是连续的并且自己坐在最左边，如果有多个的话，找最舒适的座位，如果没有连续t个，那么只给自己找个最舒适的位子，如果都满的话，输出-1.
题解：一个简单的搜索模拟，注意的是，要排序每个同学进来的时间，而且输出要按照给的顺序输出，被坑了几次，样例数据太弱了。
AC代码：
#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <cmath>#include <vector>#include <list>#include <deque>#include <queue>#include <iterator>#include <stack>#include <map>#include <set>#include <algorithm>#include <cctype>using namespace std;#define si1(a) scanf("%d",&a)#define si2(a,b) scanf("%d%d",&a,&b)#define sd1(a) scanf("%lf",&a)#define sd2(a,b) scanf("%lf%lf",&a,&b)#define ss1(s) scanf("%s",s)#define pi1(a) printf("%d\n",a)#define pi2(a,b) printf("%d %d\n",a,b)#define mset(a,b) memset(a,b,sizeof(a))#define forb(i,a,b) for(int i=a;i<b;i++)#define ford(i,a,b) for(int i=a;i<=b;i++)typedef long long LL;const int N=33;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);const double eps=1e-8;int n,m,k;int xh[N][N];struct xkn{ int time,ren; int i;}p[55];struct jilu{ int x,y;}o[55];bool cmp(xkn a,xkn b){ return a.time<b.time;}bool yes(int i,int j,int ren){ for(int k=0;k<ren;k++) if(xh[i][j+k]==-1) return false; return true;}void xiaohao(int ren,int t){ int flag1=-1,flag2=-1; int Max1,Max2; int x,y; for(int i=0;i<n;i++) for(int j=0;j<=m-ren;j++) { if(yes(i,j,ren)) { if(flag1==-1) { flag1=1; Max1=xh[i][j]; x=i; y=j; } else if(Max1<xh[i][j]) { Max1=xh[i][j]; x=i; y=j; } } } if(flag1!=-1) { //pi2(x+1,y+1); o[t].x=x+1; o[t].y=y+1; for(int k=0;k<ren;k++) xh[x][y+k]=-1; return ; } forb(i,0,n) forb(j,0,m) { if(xh[i][j]!=-1) { if(flag2==-1) { flag2=1; Max2=xh[i][j]; x=i; y=j; } else if(Max2<xh[i][j]) { Max2=xh[i][j]; x=i; y=j; } } } if(flag2!=-1) { //pi2(x+1,y+1); o[t].x=x+1; o[t].y=y+1; xh[x][y]=-1; return ; } o[t].x=-1; o[t].y=-1;}int main(){// freopen("input.txt","r",stdin); while(scanf("%d%d%d",&n,&m,&k)&&(n+m+k)) { forb(i,0,n) forb(j,0,m) si1(xh[i][j]); forb(i,0,k) { int a,b; scanf("%d:%d %d",&a,&b,&p[i].ren); p[i].time=a*60+b; p[i].i=i; } sort(p,p+k,cmp); forb(i,0,k) { xiaohao(p[i].ren,p[i].i); } forb(i,0,k)//必须按照开始的顺序输出 { if(o[i].x==-1) puts("-1"); else pi2(o[i].x,o[i].y); } } return 0;}