[2016校赛]两个简单的小题

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 10005int 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 0x3f3f3f3fint 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));    }}

结语

这两天，ACM/ICPC世界总决赛在泰国清迈进行，想来看算法也快一年了。而今年又有一所中国的大学第一次进入了world final。我想，好好努力吧，未来谁知道呢？不说了，要看直播去了。再次感谢我的队友，因为你们我不再羡慕任何人，任何事。