[2016校赛]两个简单的小题

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本文分享了作者在校内竞赛中解决两道题目的过程，包括测量药物剂量的背包问题和森林路径选择问题，并附上了C语言实现的源代码。

又是一年的校赛的季节，今年的校赛的题目感觉上还是不是那么的难，我和队友一起4个小时完成了所有的题目。总之感谢他们，在做题的时候提出了很多有益的建议和测试数据。虽然最后因为罚时只是位居第二，但是，我为你们感到骄傲！！

第一题

题目描述

Now you are asked to measure a dose of medicine with a balance and a number of weights. Certainly it is not always achievable. So you should find out the qualities which cannot be measured from the range [1,S]. S is the total quality of all the weights.

解题报告

这道题实际上就是背包问题的一个翻版，在天平上，砝码既可以放在左边，也可以放在右边，就是一个两次做循环的双肩背包。只要按照要求写就可以了, 算法复杂度为O(n*sum).

source code

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

#define MAXN 105
#define MAXM 10005

int weight[MAXN];
int n;
int dp[MAXM];
int sum;

void Input()
{
    sum = 0;
    int i;
    for(i = 1; i <= n; i++){
        scanf("%d",&weight[i]);
        sum += weight[i];
    }
}

void Solve()
{
    int i,j;
    int ans = 0;
    memset(dp,0,sizeof(dp));
    dp[0] = 1;
    for(i=1;i<=n;i++){
        for(j=sum;j>=weight[i];j--){
            if(dp[j-weight[i]]){
                dp[j] = 1;
            }
        }
        for(j=0;j<=sum;j++){
            if(dp[j]){
                dp[abs(j-weight[i])] = 1;
            }
        }
    }

    for(i=1;i<=sum;i++){
        if(!dp[i])ans++;
    }
    printf("%d\n",ans);
    if(ans){
        for(i=1;i<=sum;i++){
            if(!dp[i]){
                printf("%d",i);
                if(--ans)printf(" ");
            }
        }
        printf("\n");
    }
    return;
}

int main()
{
    while(scanf("%d",&n)!=EOF){
        Input();
        Solve();
    }
    return 0;
}

第二题

题目描述

Jimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax after a hard day, he likes to walk home. To make things even nicer, his office is on one side of a forest, and his house is on the other. A nice walk through the forest, seeing the birds and chipmunks is quite enjoyable.
The forest is beautiful, and Jimmy wants to take a different route everyday. He also wants to get home before dark, so he always takes a path to make progress towards his house. He considers taking a path from A to B to be progress if there exists a route from B to his home that is shorter than any possible route from A. Calculate how many different routes through the forest Jimmy might take.

解题报告

本题的一大重要的条件是，Jimmy只会从A到B当且仅当B存在一条回家的路比所有从A出发回家的路都要短。所以我们去求从家出发的单源最短路径，然后使用记忆化搜索就好了。

source code

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<stdbool.h>

#define MAXN 1005
#define INF 0x3f3f3f3f

int graph[MAXN][MAXN];
int n,m;
int dist[MAXN];
int total[MAXN];

void Input()
{
    int i,j,k,t;
    memset(graph,INF,sizeof(graph));
    memset(total,-1,sizeof(total));
    total[2] = 1;
    for(i=1;i<=n;i++)
        graph[i][i] = 0;

    for(i=1;i<=m;i++){
        scanf("%d %d %d",&j,&k,&t);
        graph[j][k] = graph[k][j] = t;
    }
}

void Minpath()
{
    bool visited[MAXN];
    int i,j;
    for(i=1;i<=n;i++)dist[i] = graph[2][i];

    memset(visited,false,sizeof(visited));
    visited[2] = true;

    while(1){
        int temp1 = 2;
        int temp2 = INF;
        for(j=1;j<=n;j++){
            if(!visited[j] && dist[j]<temp2){
                temp2 = dist[j];
                temp1 = j;
            }
        }
        if(visited[temp1])break;
        visited[temp1] = true;
        for(j=1;j<=n;j++){
            if(temp2+graph[temp1][j]<dist[j]){
                dist[j] = graph[temp1][j]+temp2;
            }
        }
    }
    //for(i=1;i<=n;i++)printf("%d\n",dist[i]);
}

int Dfs(int index)
{
    int i,sum=0;
    if(total[index]>=0)return total[index];
    for(i=1;i<=n;i++){
        if(graph[index][i]<INF&&dist[index]>dist[i]&&i!=index)sum+=Dfs(i);
    }
    return total[index]=sum;
}

int main()
{
    freopen("input","r",stdin);
    while(1){
        scanf("%d",&n);
        if(!n)break;
        scanf("%d",&m);
        Input();
        Minpath();
        printf("%d\n",Dfs(1));
    }
}