11 | 10月 | 2015 | 超客的水墨烟雨

poj3723

maksyuki 发表于 oj 分类，标签: 图论-最小生成树

11 10月 2015

Conscription

Windy has a country, and he wants to build an army to protect his country. He has picked up N girls and M boys and wants to collect them to be his soldiers. To collect a soldier without any privilege, he must pay 10000 RMB. There are some relationships between girls and boys and Windy can use these relationships to reduce his cost. If girl x and boy y have a relationship dand one of them has been collected, Windy can collect the other one with 10000-d RMB. Now given all the relationships between girls and boys, your assignment is to find the least amount of money Windy has to pay. Notice that only one relationship can be used when collecting one soldier.

Input

The first line of input is the number of test case.
The first line of each test case contains three integers, N, M and R.
Then R lines followed, each contains three integers x_i, y_i and d_i.
There is a blank line before each test case.

1 ≤ N, M ≤ 10000
0 ≤ R ≤ 50,000
0 ≤ x_i < N
0 ≤ y_i < M
0 < d_i < 10000

Output

For each test case output the answer in a single line.

Sample Input

5 5 8
4 3 6831
1 3 4583
0 0 6592
0 1 3063
3 3 4975
1 3 2049
4 2 2104
2 2 781

5 5 10
2 4 9820
3 2 6236
3 1 8864
2 4 8326
2 0 5156
2 0 1463
4 1 2439
0 4 4373
3 4 8889
2 4 3133

Sample Output

71071

54223

Source

POJ Monthly Contest – 2009.04.05, windy7926778

题目类型：MST

算法分析：看上去需要使用二分图来求，事实上直接将每个边权取负，求解一个MST即可

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 5e5 + 66;

struct node{ int u, v, w;};

bool operator < (const node &a, const node &b) {return a.w < b.w;}

node aa[maxn];
int parent[maxn], n1, n2, m, t;

int UnFind (int val)
{
    if (parent[val] == val) return val;
    else return parent[val] = UnFind (parent[val]);
}

int PP ()
{
    int re = n1 + n2, res = 0;
    for (int i = 1; i <= m && re > 1; i++)
    {
        if (UnFind (aa[i].u) != UnFind(aa[i].v))
        {
            parent[UnFind(aa[i].u)] = UnFind (aa[i].v);
            res += aa[i].w;
            re--;
        }
    }
    return res;
}


int main()
{
//    CFF;
    scanf ("%d", &t);
    while (t--)
    {
        for (int i = 0; i < maxn; i++) parent[i] = i;

        scanf ("%d %d %d", &n1, &n2, &m);
        for (int i = 1; i <= m; i++)
        {
            scanf ("%d %d %d", &aa[i].u, &aa[i].v, &aa[i].w);
            aa[i].v += n1;
            aa[i].w = -aa[i].w;
        }

        sort (aa + 1, aa + 1 + m);
        printf ("%d\n", 10000 * (n1 + n2) + PP ());
    }
    return 0;
}

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 5e5 + 66;

struct node{ int u, v, w;};

bool operator < (const node &a, const node &b) {return a.w < b.w;}

node aa[maxn];

int parent[maxn], n1, n2, m, t;

int UnFind (int val)

{

if (parent[val] == val) return val;

else return parent[val] = UnFind (parent[val]);

}

int PP ()

{

int re = n1 + n2, res = 0;

for (int i = 1; i <= m && re > 1; i++)

{

if (UnFind (aa[i].u) != UnFind(aa[i].v))

{

parent[UnFind(aa[i].u)] = UnFind (aa[i].v);

res += aa[i].w;

re--;

}

return res;

}

int main()

{

// CFF;

scanf ("%d", &t);

while (t--)

{

for (int i = 0; i < maxn; i++) parent[i] = i;

scanf ("%d %d %d", &n1, &n2, &m);

for (int i = 1; i <= m; i++)

{

scanf ("%d %d %d", &aa[i].u, &aa[i].v, &aa[i].w);

aa[i].v += n1;

aa[i].w = -aa[i].w;

}

sort (aa + 1, aa + 1 + m);

printf ("%d\n", 10000 * (n1 + n2) + PP ());

}

return 0;

}

poj3255

maksyuki 发表于 oj 分类，标签: 图论-次\k短路

11 10月 2015

Roadblocks

Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path.

The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1..N. Bessie starts at intersection 1, and her friend (the destination) is at intersection N.

The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path).

Input

Line 1: Two space-separated integers: N and R
Lines 2..R+1: Each line contains three space-separated integers: A, B, and D that describe a road that connects intersections A and B and has length D (1 ≤ D ≤ 5000)

Output

Line 1: The length of the second shortest path between node 1 and node N

Sample Input

4 4
1 2 100
2 4 200
2 3 250
3 4 100

Sample Output

450

Hint

Two routes: 1 -> 2 -> 4 (length 100+200=300) and 1 -> 2 -> 3 -> 4 (length 100+250+100=450)

Source

USACO 2006 November Gold

题目类型：次短路

算法分析：直接使用Dijkstra算法同时维护最短路和次短路，若能够更新最短路则先更新最短路，然后再更新次短路

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 5e3 + 66;

vector <PII> edge[maxn];
int n, dis[maxn][2];//n表示节点的个数，dis[i][0]表示从源点到i的最短路，dis[i][1]表示从源点到i的次短路

void Dijkstra (int s)
{
    for (int i = 0; i < maxn; i++)
        for (int j = 0; j < 2; j++)
            dis[i][j] = INF;

    dis[s][0] = 0;//源点到自己的最短路是0，十分重要!!!
    priority_queue <PII, vector <PII>, greater <PII> > qu;
    qu.push (PII (0, s));
    while (!qu.empty ())
    {
        PII ta = qu.top (); qu.pop ();
        int v1 = ta.second, d1 = ta.first;
        if (dis[v1][1] < d1) continue;
        for (int i = 0; i < edge[v1].size (); i++)
        {
            PII tb = edge[v1][i];
            int v2 = tb.first, w2 = tb.second;
            int d2 = d1 + w2;
            if (dis[v2][0] > d2)
            {
                swap (dis[v2][0], d2);
                qu.push (PII (dis[v2][0], v2));
            }

            if (dis[v2][0] < d2 && dis[v2][1] > d2)
            {
                dis[v2][1] = d2;
                qu.push (PII (dis[v2][1], v2));
            }
        }
    }
}


int main()
{
//    CFF;
    int m;
    while (scanf ("%d %d", &n, &m) != EOF)
    {
        for (int i = 0; i < maxn; i++) edge[i].clear ();

        for (int i = 1; i <= m; i++)
        {
            int u, v, w;
            scanf ("%d %d %d", &u, &v, &w);
            edge[u].PB (PII (v, w));
            edge[v].PB (PII (u, w));
        }

        Dijkstra (1);
        printf ("%d\n", dis[n][1]);
    }
    return 0;
}

100

101

102

103

104

105

106

107

108

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 5e3 + 66;

vector <PII> edge[maxn];

int n, dis[maxn][2];//n表示节点的个数，dis[i][0]表示从源点到i的最短路，dis[i][1]表示从源点到i的次短路

void Dijkstra (int s)

{

for (int i = 0; i < maxn; i++)

for (int j = 0; j < 2; j++)

dis[i][j] = INF;

dis[s][0] = 0;//源点到自己的最短路是0，十分重要!!!

priority_queue <PII, vector <PII>, greater <PII> > qu;

qu.push (PII (0, s));

while (!qu.empty ())

{

PII ta = qu.top (); qu.pop ();

int v1 = ta.second, d1 = ta.first;

if (dis[v1][1] < d1) continue;

for (int i = 0; i < edge[v1].size (); i++)

{

PII tb = edge[v1][i];

int v2 = tb.first, w2 = tb.second;

int d2 = d1 + w2;

if (dis[v2][0] > d2)

{

swap (dis[v2][0], d2);

qu.push (PII (dis[v2][0], v2));

}

if (dis[v2][0] < d2 && dis[v2][1] > d2)

{

dis[v2][1] = d2;

qu.push (PII (dis[v2][1], v2));

}

int main()

{

// CFF;

int m;

while (scanf ("%d %d", &n, &m) != EOF)

{

for (int i = 0; i < maxn; i++) edge[i].clear ();

for (int i = 1; i <= m; i++)

{

int u, v, w;

scanf ("%d %d %d", &u, &v, &w);

edge[u].PB (PII (v, w));

edge[v].PB (PII (u, w));

}

Dijkstra (1);

printf ("%d\n", dis[n][1]);

}

return 0;

}

超客的水墨烟雨

poj3723

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