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2014 Super Training #10 C Shadow --SPFA/随便搞/DFS

原题: FZU 2169 http://acm.fzu.edu.cn/problem.php?pid=2169

这题貌似有两种解法,DFS和SPFA,但是DFS怎么都RE,SPFA也要用邻接表表示边,用向量表示的话会TLE,而且用SPFA有一个异或,就是题目说要沿最短路走到都城,但是SPFA是走最短路去消灭叛军,然后再走回都城,我不知道怎么回事,不知道能不能有大神解释。因为这样的话,有多少叛军就能消灭多少叛军了,那就不要用什么算法 了,直接一个统计。于是试了一下,居然A了,瞬间变成大水题,我无法再评价这个题目了,就当消遣了。

SPFA法:依次从每个军队城市出发做一次SPFA,看到有能到的(肯定能到啊)叛军就将其消灭。

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <queue>
#define Mod 1000000007
using namespace std;
#define N 100007

struct Edge
{
    int v,next;
}G[2*N];
int head[N],tot;
int army[N],rebel[N];
int vis[N],dis[N];
int res;
int n,m;

void addedge(int u,int v)
{
    G[tot].v = v;
    G[tot].next = head[u];
    head[u] = tot++;
}

void SPFA(int s)
{
    int i;
    memset(vis,0,sizeof(vis));
    queue<int> que;
    while(!que.empty())
        que.pop();
    que.push(s);
    vis[s] = 1;
    dis[s] = 0;
    while(!que.empty())
    {
        int tmp = que.front();
        que.pop();
        vis[tmp] = 0;
        for(i=head[tmp];i!=-1;i=G[i].next)
        {
            int v = G[i].v;
            if(dis[v] > dis[tmp] + 1)
            {
                dis[v] = dis[tmp]+1;
                if(!vis[v])
                {
                    que.push(v);
                    vis[v] = 1;
                }
            }
        }
    }
}

int main()
{
    int i,j,x;
    int u,v;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(army,0,sizeof(army));
        memset(rebel,0,sizeof(rebel));
        memset(head,-1,sizeof(head));
        tot = 0;
        int cnt = 0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&rebel[i]);
            if(rebel[i])
                cnt++;
        }
        for(i=1;i<=m;i++)
            scanf("%d",&army[i]);
        for(i=0;i<n-1;i++)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        res = 0;
        for(i=1;i<=m;i++)
        {
            SPFA(army[i]);
            for(j=1;j<=n;j++)
            {
                if(dis[j] != Mod)
                {
                    if(rebel[j])
                    {
                        res += rebel[j];
                        rebel[j] = 0;
                        cnt--;
                    }
                }
            }
            if(cnt == 0)  //已经消灭完
                break;
        }
        printf("%d\n",res);
    }
    return 0;
}
View Code

 

DFS法(选GNU C++ 会Runtime Error,要选Visual C++):

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define Mod 1000000007
#define ll long long
using namespace std;
#define N 100007

struct Edge
{
    int v,next;
}G[2*N];
int head[N],tot;
int army[N],rebel[N];
ll sum;
int n,m;

void addedge(int u,int v)
{
    G[tot].v = v;
    G[tot].next = head[u];
    head[u] = tot++;
}

int dfs(int u,int val,int fa)
{
    int soni = 0;
    if(army[u])  //找到军队
    {
        sum += val;
        soni++;
    }
    for(int i=head[u];i!=-1;i=G[i].next)
    {
        int v = G[i].v;
        if(v == fa)
            continue;
        soni += dfs(v,val+rebel[u],u);
    }
    if(soni > 1)
        sum -= (soni-1)*rebel[u];
    return soni;
}

int main()
{
    int i,j,x;
    int u,v;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(army,0,sizeof(army));
        memset(rebel,0,sizeof(rebel));
        memset(head,-1,sizeof(head));
        tot = 0;
        for(i=1;i<=n;i++)
            scanf("%d",&rebel[i]);
        for(i=1;i<=m;i++)
        {
            scanf("%d",&x);
            army[x] = 1;
        }
        for(i=0;i<n-1;i++)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        sum = 0;
        dfs(1,0,-1);
        printf("%lld\n",sum);
    }
    return 0;
}
View Code

 

直接统计:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
#define N 100007

int army[N],rebel[N];
int res;
int n,m;

int main()
{
    int i,j,x;
    int u,v;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(rebel,0,sizeof(rebel));
        int cnt = 0;
        res = 0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&rebel[i]);
            if(rebel[i])
                res += rebel[i];
        }
        for(i=1;i<=m;i++)
            scanf("%d",&army[i]);
        for(i=0;i<n-1;i++)
            scanf("%d%d",&u,&v);
        if(m == 0)
        {
            puts("0");
            continue;
        }
        printf("%d\n",res);
    }
    return 0;
}
View Code

 

posted @ 2014-07-08 19:40  whatbeg  阅读(242)  评论(0编辑  收藏  举报