HDOJ 1224 记忆搜索

前两天才做了一道记忆搜索题，所以这道题很快就有思路了。

如果一个路径，它经过的城市的interesting point之和最大，但是这条路径的终点不是起点，则这条路径不能取。

#include <iostream>
using namespace std;

const int MAX_CITY = 105;
int map[MAX_CITY + 1][MAX_CITY + 1];/* 最后一个点是起点 */
int ans[MAX_CITY + 1],point[MAX_CITY + 1],nextCity[MAX_CITY + 1];

int DFS(int n,int cityNum)/* n是当前所在城市 */
{
    int maxPoint = 0;/* 从n的下一点i出发，所能积累的最大point值 */
    int isReturnable = 0;/* 是否能返回出发点 */
    for (int i = (n + 1);i <= (cityNum + 1);i ++)
    {
        if (map[n][i] == 1)
        {
            int temp = 0;
            if (ans[i] == 0)/* 第i个城市的值没有计算过，先计算 */
                temp = DFS(i,cityNum);
            else
                temp = ans[i];
            if (temp > maxPoint)
            {
                maxPoint = temp;
                nextCity[n] = i;
            }
            if (temp != 0 || (i == (cityNum + 1)))
                isReturnable = 1;
        }
    }
    if (isReturnable == 1)
        ans[n] = point[n] + maxPoint;
    return ans[n];
}

int main ()
{
    int caseNum = 1;
    scanf("%d",&caseNum);
    for (int caseCount = 1;caseCount <= caseNum;caseCount ++)
    {
        memset(map,0,sizeof(map));
        memset(ans,0,sizeof(ans));
        memset(nextCity,0,sizeof(nextCity));
        int cityNum,flightNum;
        scanf("%d",&cityNum);
        for (int i = 1;i <= cityNum;i ++)
            scanf("%d",&point[i]);
        point[0] = point[cityNum + 1] = 0;
        scanf("%d",&flightNum);
        for (int i = 1;i <= flightNum;i ++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            map[a][b] = 1;
        }
         printf("CASE %d#\n",caseCount);
        printf("points : %d\n",DFS(1,cityNum));
        printf("circuit : ");
        int curCity = 1;
        while (curCity != 0)
        {
            printf("%d->",curCity);
            curCity = nextCity[curCity];
        }
        printf("1\n");
        if (caseCount < caseNum)
            printf("\n");
    }
    return 0;
}