USACO Section 1.4 Packing Rectangles （POJ 1169）搜索

这是一个很蛋疼的搜索题，虽然题中只给了5种矩形组合方式，但是搜索时的长和宽并不是唯一的，所以就产生了很多种组合方式.需要注意的是第4种和第3种图形实际上能规划成一种。 而产生这些组合最好用的应该就是DFS了，而我当时不想动脑子，使用的就是纯枚举，巨大的代码量，最终好歹也能过了。

/*ID: sdj22251PROG: packrecLANG: C++*/#include <iostream>#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <cctype>#include <string>#include <cstring>#include <cmath>#include <ctime>#define MAX 2000000000#define M 1007#define LOCALusing namespace std;struct wwj{    int x, y;}a[4], ans[21050];bool v[5];wwj f1(wwj a, wwj b, wwj c, wwj d){    wwj rec;    rec.x = a.x + b.x + c.x + d.x;    rec.y = max(max(a.y, b.y), max(c.y, d.y));    return rec;}wwj f2(wwj a, wwj b, wwj c, wwj d){    wwj rec;    rec.x = max(a.x, b.x + c.x + d.x);    rec.y = a.y + max(d.y, max(b.y, c.y));    return rec;}wwj f3(wwj a, wwj b, wwj c, wwj d){    wwj rec;    rec.x = max(a.x+b.x, c.x)+d.x;rec.y = max(max(a.y, b.y)+c.y, d.y);    return rec;}wwj f4(wwj a, wwj b, wwj c, wwj d){    wwj rec;    rec.x = a.x + max(b.x, c.x) + d.x;rec.y = max(max(a.y, b.y + c.y), d.y);    return rec;}wwj f6(wwj a, wwj b, wwj c, wwj d){    wwj rec;    rec.x = a.x+b.x;rec.y = max(a.y+c.y, b.y+d.y);if (a.y < b.y)rec.x = max(rec.x, c.x+b.x);if (a.y+c.y > b.y)rec.x = max(rec.x, c.x+d.x);if (b.y < a.y)rec.x = max(rec.x, a.x+d.x);rec.x = max(rec.x, c.x);rec.x = max(rec.x, d.x);    return rec;}bool cmp(wwj a, wwj b){    if(a.x * a.y == b.x * b.y)    return a.x < b.x;    else return a.x * a.y < b.x * b.y;}int main(){#ifdef LOCAL    freopen("packrec.in","r",stdin);    freopen("packrec.out","w",stdout);#endif    int i, j, k, q;    int count = 0;    for(i = 0; i < 4; i++)    scanf("%d%d", &a[i].x, &a[i].y);    for(i = 0; i < 4; i++)    {        v[i] = true;        for(j = 0; j < 4; j++)        {            if(v[j] == true)            continue;            v[j] = true;            for(k = 0; k < 4; k++)            {                if(v[k] == true)                continue;                v[k] = true;                for(q = 0; q < 4; q++)                {                    if(v[q] == true)                    continue;                    wwj fk1, fk2, fk3, fk4;                    fk1.x = a[i].y;                    fk1.y = a[i].x;                    fk2.x = a[j].y;                    fk2.y = a[j].x;                    fk3.x = a[k].y;                    fk3.y = a[k].x;                    fk4.x = a[q].y;                    fk4.y = a[q].x;                    ans[count++] = f1(a[i], a[j], a[k], a[q]);                    ans[count++] = f2(a[i], a[j], a[k], a[q]);                    ans[count++] = f3(a[i], a[j], a[k], a[q]);                    ans[count++] = f4(a[i], a[j], a[k], a[q]);                    ans[count++] = f6(a[i], a[j], a[k], a[q]);                    ans[count++] = f1(fk1, a[j], a[k], a[q]);                    ans[count++] = f2(fk1, a[j], a[k], a[q]);                    ans[count++] = f3(fk1, a[j], a[k], a[q]);                    ans[count++] = f4(fk1, a[j], a[k], a[q]);                    ans[count++] = f6(fk1, a[j], a[k], a[q]);                    ans[count++] = f1(fk1, a[j], fk3, a[q]);                    ans[count++] = f2(fk1, a[j], fk3, a[q]);                    ans[count++] = f3(fk1, a[j], fk3, a[q]);                    ans[count++] = f4(fk1, a[j], fk3, a[q]);                    ans[count++] = f6(fk1, a[j], fk3, a[q]);                    ans[count++] = f1(fk1, a[j], a[k], fk4);                    ans[count++] = f2(fk1, a[j], a[k], fk4);                    ans[count++] = f3(fk1, a[j], a[k], fk4);                    ans[count++] = f4(fk1, a[j], a[k], fk4);                    ans[count++] = f6(fk1, a[j], a[k], fk4);                    ans[count++] = f1(fk1, fk2, a[k], a[q]);                    ans[count++] = f2(fk1, fk2, a[k], a[q]);                    ans[count++] = f3(fk1, fk2, a[k], a[q]);                    ans[count++] = f4(fk1, fk2, a[k], a[q]);                    ans[count++] = f6(fk1, fk2, a[k], a[q]);                    ans[count++] = f1(fk1, fk2, a[k], fk4);                    ans[count++] = f2(fk1, fk2, a[k], fk4);                    ans[count++] = f3(fk1, fk2, a[k], fk4);                    ans[count++] = f4(fk1, fk2, a[k], fk4);                    ans[count++] = f6(fk1, fk2, a[k], fk4);                    ans[count++] = f1(fk1, fk2, fk3, a[q]);                    ans[count++] = f2(fk1, fk2, fk3, a[q]);                    ans[count++] = f3(fk1, fk2, fk3, a[q]);                    ans[count++] = f4(fk1, fk2, fk3, a[q]);                    ans[count++] = f6(fk1, fk2, fk3, a[q]);                    ans[count++] = f1(fk1, fk2, fk3, fk4);                    ans[count++] = f2(fk1, fk2, fk3, fk4);                    ans[count++] = f3(fk1, fk2, fk3, fk4);                    ans[count++] = f4(fk1, fk2, fk3, fk4);                    ans[count++] = f6(fk1, fk2, fk3, fk4);                    ans[count++] = f1(a[i], fk2, a[k], a[q]);                    ans[count++] = f2(a[i], fk2, a[k], a[q]);                    ans[count++] = f3(a[i], fk2, a[k], a[q]);                    ans[count++] = f4(a[i], fk2, a[k], a[q]);                    ans[count++] = f6(a[i], fk2, a[k], a[q]);                    ans[count++] = f1(a[i], fk2, a[k], fk4);                    ans[count++] = f2(a[i], fk2, a[k], fk4);                    ans[count++] = f3(a[i], fk2, a[k], fk4);                    ans[count++] = f4(a[i], fk2, a[k], fk4);                    ans[count++] = f6(a[i], fk2, a[k], fk4);                    ans[count++] = f1(a[i], fk2, fk3, a[q]);                    ans[count++] = f2(a[i], fk2, fk3, a[q]);                    ans[count++] = f3(a[i], fk2, fk3, a[q]);                    ans[count++] = f4(a[i], fk2, fk3, a[q]);                    ans[count++] = f6(a[i], fk2, fk3, a[q]);                    ans[count++] = f1(a[i], fk2, fk3, fk4);                    ans[count++] = f2(a[i], fk2, fk3, fk4);                    ans[count++] = f3(a[i], fk2, fk3, fk4);                    ans[count++] = f4(a[i], fk2, fk3, fk4);                    ans[count++] = f6(a[i], fk2, fk3, fk4);                    ans[count++] = f1(a[i], a[j], fk3, a[q]);                    ans[count++] = f2(a[i], a[j], fk3, a[q]);                    ans[count++] = f3(a[i], a[j], fk3, a[q]);                    ans[count++] = f4(a[i], a[j], fk3, a[q]);                    ans[count++] = f6(a[i], a[j], fk3, a[q]);                    ans[count++] = f1(a[i], a[j], fk3, fk4);                    ans[count++] = f2(a[i], a[j], fk3, fk4);                    ans[count++] = f3(a[i], a[j], fk3, fk4);                    ans[count++] = f4(a[i], a[j], fk3, fk4);                    ans[count++] = f6(a[i], a[j], fk3, fk4);                    ans[count++] = f1(a[i], a[j], a[k], fk4);                    ans[count++] = f2(a[i], a[j], a[k], fk4);                    ans[count++] = f3(a[i], a[j], a[k], fk4);                    ans[count++] = f4(a[i], a[j], a[k], fk4);                    ans[count++] = f6(a[i], a[j], a[k], fk4);                }                v[k] = false;            }            v[j] = false;        }        v[i] = false;    }    sort(ans, ans + count, cmp);    printf("%d\n", ans[0].x * ans[0].y);    int t = ans[0].x * ans[0].y;    for(i = 0; i < count; i++)    {        if(ans[i].x * ans[i].y == t)        {            ans[i].x = min(ans[i].x, ans[i].y);            ans[i].y = t / ans[i].x;        }    }    sort(ans, ans + count, cmp);    for(i = 0; i < count; i++)    {        if(ans[i].x * ans[i].y == ans[0].x * ans[0].y)        {            if(i >= 1 && ans[i].x == ans[i - 1].x)            continue;            printf("%d %d\n", ans[i].x, ans[i].y);        }    }    return 0;}

附官方题解一份，代码风格让我觉得很飘渺，怎么也看的不是很懂

This program is straightforward, but a bit long due to the geometry involved.

There are 24 permutations of the 4 rectangles, and for each permutation, 16 different ways to orient them. 

We generate all such orientations of permutations, and put the blocks together in each of the 6 different ways, recording the smallest rectangles we find.

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <assert.h>typedef struct Rect Rect;struct Rect {    int wid;    int ht;};Rectrotate(Rect r){    Rect nr;    nr.wid = r.ht;    nr.ht = r.wid;    return nr;}intmax(int a, int b){    return a > b ? a : b;}intmin(int a, int b){    return a < b ? a : b;}int tot;int bestarea;int bestht[101];voidrecord(Rect r){    int i;    if(r.wid*r.ht < tot)        *(long*)0=0;    if(r.wid*r.ht < bestarea || bestarea == 0) {        bestarea = r.wid*r.ht;        for(i=0; i<=100; i++)            bestht[i] = 0;    }    if(r.wid*r.ht == bestarea)        bestht[min(r.wid, r.ht)] = 1;}voidcheck(Rect *r){    Rect big;    int i;    /* schema 1: all lined up next to each other */    big.wid = 0;    big.ht = 0;    for(i=0; i<4; i++) {        big.wid += r[i].wid;        big.ht = max(big.ht, r[i].ht);    }    record(big);    /* schema 2: first three lined up, fourth on bottom */    big.wid = 0;    big.ht = 0;    for(i=0; i<3; i++) {        big.wid += r[i].wid;        big.ht = max(big.ht, r[i].ht);    }    big.ht += r[3].ht;    big.wid = max(big.wid, r[3].wid);    record(big);    /* schema 3: first two lined up, third under them, fourth to side */    big.wid = r[0].wid + r[1].wid;    big.ht = max(r[0].ht, r[1].ht);    big.ht += r[2].ht;    big.wid = max(big.wid, r[2].wid);    big.wid += r[3].wid;    big.ht = max(big.ht, r[3].ht);    record(big);    /* schema 4, 5: first two rectangles lined up, next two stacked */    big.wid = r[0].wid + r[1].wid;    big.ht = max(r[0].ht, r[1].ht);    big.wid += max(r[2].wid, r[3].wid);    big.ht = max(big.ht, r[2].ht+r[3].ht);    record(big);    /*     * schema 6: first two pressed next to each other, next two on top, like:      * 2 3     * 0 1     */    big.ht = max(r[0].ht+r[2].ht, r[1].ht+r[3].ht);    big.wid = r[0].wid + r[1].wid;    /* do 2 and 1 touch? */    if(r[0].ht < r[1].ht)        big.wid = max(big.wid, r[2].wid+r[1].wid);    /* do 2 and 3 touch? */    if(r[0].ht+r[2].ht > r[1].ht)        big.wid = max(big.wid, r[2].wid+r[3].wid);    /* do 0 and 3 touch? */    if(r[1].ht < r[0].ht)        big.wid = max(big.wid, r[0].wid+r[3].wid);    /* maybe 2 or 3 sits by itself */    big.wid = max(big.wid, r[2].wid);    big.wid = max(big.wid, r[3].wid);    record(big);    }voidcheckrotate(Rect *r, int n){    if(n == 4) {        check(r);        return;    }    checkrotate(r, n+1);    r[n] = rotate(r[n]);    checkrotate(r, n+1);    r[n] = rotate(r[n]);}voidcheckpermute(Rect *r, int n){    Rect t;    int i;    if(n == 4)        checkrotate(r, 0);    for(i=n; i<4; i++) {        t = r[n], r[n] = r[i], r[i] = t;    /* swap r[i], r[n] */        checkpermute(r, n+1);        t = r[n], r[n] = r[i], r[i] = t;    /* swap r[i], r[n] */    }}voidmain(void){    FILE *fin, *fout;    Rect r[4];    int i;    fin = fopen("packrec.in", "r");    fout = fopen("packrec.out", "w");    assert(fin != NULL && fout != NULL);    for(i=0; i<4; i++)        fscanf(fin, "%d %d", &r[i].wid, &r[i].ht);    tot=(r[0].wid*r[0].ht+r[1].wid*r[1].ht+r[2].wid*r[2].ht+r[3].wid*r[3].ht);    checkpermute(r, 0);    fprintf(fout, "%d\n", bestarea);    for(i=0; i<=100; i++)        if(bestht[i])            fprintf(fout, "%d %d\n", i, bestarea/i);    exit(0);}