Steal the Treasure

Time Limit: 10000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 775    Accepted Submission(s): 213

Problem Description The alliance of thieves decides to steal the treasure from country A. There are n cities in country A. Cities are connected by directional or bidirectional road. To avoid the risk, the king of country A divides his treasure and hides them in some place on the road.   The alliance has found out the secret of the king. They get a map of country A which shows the location and the quantity of treasure on each road. In order to make the maximum profit and reduce the least loss, the alliance determines to send n thieves respectively to each city (one city one thief). At the appointed time, each thief chooses one road (if there is a road and notice that the road may have direction) to get to its corresponding city. Then he can steal the treasure on that road. After stealing, all the thieves return back to their base immediately.   The heads of the alliance wonder to know the quantity of the treasure they can steal at most. Input There are multiple cases. Input is terminated by EOF.   For each case, the first line contains two integers n (1<=n<=1000) and m (0<=m<=n*(n-1)/2), representing the number of cities and the number of roads in country A. The following m lines, each line contains four integers x, y (1<=x, y<=n, x≠y), d (0<=d<=1), w (0<=w<=1000), which means that there is a road from city x to city y, d=0 shows this road is bidirectional and d=1 shows it is directional and x the starting point, w is the quantity of treasure on the road.   We guarantee that the road (x, y) and (y, x) will never appear together in the same case. Output For each case, output the maximum quantity of treasure the alliance can get. Sample Input 2 1 1 2 0 10 5 5 1 2 1 0 1 3 1 10 2 3 0 20 3 4 0 30 4 2 1 40 Sample Output 10 100  題目大意：有n個城市，這些城市由m條道路連通，每一條道路都有著一定的權值（財富），這些道路有的是可以雙向連通的，有的是單向的，只能從一個點出發，現在每一個城市都有著一個小偷，在特定時刻，這些小偷可以從這個城市前往任一條道路（如果可以），並拿走這條道路上的財富，現在問你最多能拿到多少財富。 思路分析：首先肯定是貪心，將邊按照邊權從大到小排序，到底能不能拿這個路上的財富是由他的端點城市決定的，對於有向邊，如果起點未被標記，那麼就拿走財富，標記起點，對於無向邊，如果有一個點沒被標記，就標記那個點，否則就用並查集將其縮為一點。 程式碼：

#include <iostream>
#include
 
<cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn=1000+10;
struct node
{
    int x,y;
    int d,w;
};
node edge[maxn*maxn/2];
int fa[maxn];
int root(int x)
{
    return (x==fa[x])?x:fa[x]=root(fa[x]);
}
bool cmp(node a,node b)
{
    return
 
 a.w>b.w;
}
bool vis[maxn];
int n,m;
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(vis,false,sizeof(vis));
        for(int i=1;i<=n;i++)
            fa[i]=i;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d%d",&edge[i].x,&edge[i].y,&edge[i].d,&edge[i].w);
        }
        sort(edge,edge+m,cmp);
        int ans=0;
        for(int i=0;i<m;i++)
        {
            int fx=root(edge[i].x);
            int fy=root(edge[i].y);
            if(vis[fx]&&vis[fy]) continue;
            if(edge[i].d==1&&vis[fx]) continue;
            ans+=edge[i].w;
            if(edge[i].d==1) vis[fx]=true;
            else
            {
                if(fx==fy) vis[fx]=true;
                else if(vis[fx]) vis[fy]=true;
                else if(vis[fy]) vis[fx]=true;
                else fa[fx]=fy;//縮點
            }
        }
        printf("%d\n",ans);
    }
}