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HDU 2448 Mining Station on the Sea

2019年04月06日 ⁄ 综合 ⁄ 共 2020字 ⁄ 字号 评论关闭

预处理轮船到各个港口的最短路径,然后KM解之即可。

/*HDU 2448*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
using namespace std;

const int maxn = 310;
const int INF = 0x3f3f3f3f;

int n, m;

int W[maxn][maxn];
int Lx[maxn], Ly[maxn];

int Left[maxn];
bool S[maxn], T[maxn];

bool match(int i)
{
    S[i] = 1;
    for(int j = 1; j <= m; j++) if(Lx[i]+Ly[j] == W[i][j] && !T[j])
    {
        T[j] = 1;
        if(!Left[j] || match(Left[j]))
        {
            Left[j] = i;
            return 1;
        }
    }
    return 0;
}

void update()
{
    int a = INF;
    for(int i = 1; i <= n; i++) if(S[i])
    for(int j = 1; j <= m; j++) if(!T[j])
        a = min(a, Lx[i]+Ly[j]-W[i][j]);
        
    for(int i = 1; i <= n; i++)
    {
        if(S[i]) Lx[i] -= a;
    }
    for(int j = 1; j <= m; j++)
    {
        if(T[j]) Ly[j] += a;
    }
}

void KM()
{
    memset(Left, 0, sizeof(Left));
    memset(Lx, 0, sizeof(Lx));
    memset(Ly, 0, sizeof(Ly));
    
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
            Lx[i] = max(Lx[i], W[i][j]);
    }
    for(int i = 1; i <= n; i++)
    {
        for(;;)
        {
            for(int j = 1; j <= m; j++) S[j] = T[j] = 0;
            if(match(i)) break; else update();
        }
    }
}

inline void readint(int &x)
{
    char c;
    c = getchar();
    while(!isdigit(c)) c = getchar();
    
    x = 0;
    while(isdigit(c)) x = x*10+c-'0', c = getchar();
}

inline void writeint(int x)
{
    if(x > 9) writeint(x/10);
    putchar(x%10+'0');
}

int N, M, K, P;

int station[maxn];
int d[maxn];
int G[maxn][maxn];

void Dijkstra(int s)
{
	bool vis[maxn] = {0};
	for(int i = 1; i <= N+M; i++) d[i] = (i == s)? 0:INF;
	
	for(int i = 1; i <= N+M; i++)
	{
		int x, MIN = INF;
		for(int y = 1; y <= N+M; y++) if(!vis[y] && d[y] <= MIN) MIN = d[x=y];
		vis[x] = 1;
		for(int y = 1; y <= N+M; y++) d[y] = min(d[y], d[x] + G[x][y]);
	}
}

void read_case()
{
	n = m = N;
	memset(G, INF, sizeof(G));
	
	for(int i = 1; i <= N; i++) readint(station[i]);
	
	for(int i = 1; i <= K; i++)
	{
		int x, y, z;
		readint(x), readint(y), readint(z);
		if(z < G[x][y])
			G[x][y] = G[y][x] = z;
	}
	
	for(int i = 1; i <= P; i++)
	{
		int x, y, z;
		readint(x), readint(y), readint(z);
		if(z < G[y][x+M])
			G[y][x+M] = z;
	}
}

void build()
{
	for(int i = 1; i <= N; i++)
	for(int j = 1; j <= N; j++) W[i][j] = -INF;
	
	for(int i = 1; i <= N; i++)
	{
		Dijkstra(station[i]);
		for(int j = M+1; j <= N+M; j++)
		{
			if(d[j] != INF) W[i][j-M] = -d[j];
		}
	}
}

void solve()
{
	read_case();
	build();
	
	KM();
	int ans = 0;
	for(int i = 1; i <= m; i++) if(Left[i] && W[Left[i]][i] != -INF) ans -= W[Left[i]][i];
	printf("%d\n", ans);
}

int main()
{
	while(~scanf("%d%d%d%d", &N, &M, &K, &P))
	{
		solve();
	}
	return 0;
}

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