POJ - 1010 STAMPS

       指定m种“总面值”，对每种“总面值”，求解满足如下条件的组合以达到该“总面值”

（1）       所用邮票在n种中可以重复选取

（2）       所用邮票张数〈＝4

（3）       尽量多的使用那个不同种类的邮票 Max (Stamp Types)

（4）       若有多种方案满足（3），则选取张数最小的一种方案 Min (Stamp Num)

（5）       若有多种方案满足(3)(4),则选取“最大面额”最高的一种方案。 Max(Heightest Value)

（6）       若有多种方案满足（3）（4）（5） 则输出 “tie”

思路：经典的搜索题目：超过4张的按4张处理，但看了大牛的说5张才是合适的，其次就是4层的搜索，还有就是当我们排序处理后，可以更快的处理need

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 30;

int val[MAXN],pv,Time[MAXN],solve[6];
int flag,flag_tie;
int bestSolve[6];

int max4(int s[]){
	int a = max(s[1], s[2]);
	int b = max(s[3], s[4]);
	return max(a, b);
}

void best(int num, int type){
	bestSolve[0] = num;
	bestSolve[5] = type;
	for (int i = 1; i <= 4; i++)
		bestSolve[i] = solve[i];
	return;
}

void dfs(int need, int num, int type, int pre){
	if (num == 5)
		return;
	if (need == 0){
		if (!flag){
			if (type == bestSolve[5]){
				if (num == bestSolve[0]){
					int Maxs = max4(solve);
					int Maxbs = max4(bestSolve);
					if (Maxs == Maxbs)
						flag_tie = 1;
					else if (Maxs > Maxbs){
						flag_tie = 0;
						best(num, type);
					}
				}
				else if (num < bestSolve[0]){
					flag_tie = 0;
					best(num, type);
				}
			}
			else if (type > bestSolve[5]){
				flag_tie = 0;
				best(num, type);
			}
		}
		else {
			flag = 1;
			best(num, type);
		}
		return;
	}
	for (int i = pre; i < pv; i++){
		if (need >= val[i]){
			solve[num+1] = val[i];
			if (Time[i]){
				Time[i]++;
				dfs(need-val[i], num+1, type, i);
			}
			else {
				Time[i]++;
				dfs(need-val[i], num+1, type+1, i);
			}
			solve[num+1] = 0;
			Time[i]--;
		}
		else return;
	}
	return; 
}

int main(){
	while (1){
		pv = 0;
		int type[MAXN] = {0};
		int tmp;
		while (1){
			if (scanf("%d", &tmp) == EOF)
				exit(0);
			if (tmp == 0)
				break;
			if (type[tmp] < 4){
				type[tmp]++;
				val[pv++] = tmp;
			}
		}
		sort(val, val+pv);
		int need;
		while (scanf("%d", &need) != EOF && need){
			flag = 0;
			flag_tie = 0;
			memset(solve, 0, sizeof(solve));
			memset(bestSolve, 0, sizeof(bestSolve));
			memset(Time, 0, sizeof(Time));
			dfs(need, 0, 0, 0);
			printf("%d", need);
			if (bestSolve[0] == 0)
				printf(" ---- none\n");
			else {
				printf(" (%d):", bestSolve[5]);
				if (flag_tie)
					printf(" tie\n");
				else {
					sort(bestSolve+1, bestSolve+5);
					for (int i = 1; i <= 4; i++){
						if (bestSolve[i] == 0)
							continue;
						printf(" %d", bestSolve[i]);
					}
					printf("\n");
				}
			}
		}
	}														
	return 0;
}

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POJ - 1010 STAMPS

原文：http://blog.csdn.net/u011345136/article/details/26358075