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  • HDU 2448 Mining Station on the Sea

    HDU_2448

        首先求出各个港口和船之间的最短路,然后用KM算法求二分图最优匹配即可。

    #include<stdio.h>
    #include<string.h>
    #define MAXD 110
    #define INF 1000000000
    #define MAX 100000
    int M, K, P, N, d[MAXD], G[MAXD][MAXD], yM[MAXD];
    int A[MAXD], B[MAXD], slack, visx[MAXD], visy[MAXD];
    int f[2 * MAXD][2 * MAXD];
    void init()
    {
    int i, j, k, a, b, c, temp;
    memset(f, -1, sizeof(f));
    for(i = 0; i < N; i ++)
    {
    scanf("%d", &d[i]);
    d[i] --;
    }
    for(i = 0; i < K; i ++)
    {
    scanf("%d%d%d", &a, &b, &c);
    a --;
    b --;
    f[a][b] = f[b][a] = c;
    }
    for(i = 0; i < M; i ++)
    f[i][i] = 0;
    for(k = 0; k < M; k ++)
    for(i = 0; i < M; i ++)
    for(j = 0; j < M; j ++)
    if(f[i][k] != -1 && f[k][j] != -1)
    {
    temp = f[i][k] + f[k][j];
    if(f[i][j] == -1 || temp < f[i][j])
    f[i][j] = temp;
    }
    memset(G, 0, sizeof(G));
    for(i = 0; i < P; i ++)
    {
    scanf("%d%d%d", &a, &b, &c);
    a --;
    b --;
    for(j = 0; j < N; j ++)
    {
    temp = MAX - (c + f[b][d[j]]);
    if(temp > G[a][j])
    G[a][j] = temp;
    }
    }
    }
    int searchpath(int u)
    {
    int v, temp;
    visx[u] = 1;
    for(v = 0; v < N; v ++)
    if(!visy[v])
    {
    temp = A[u] + B[v] - G[u][v];
    if(temp == 0)
    {
    visy[v] = 1;
    if(yM[v] == -1 || searchpath(yM[v]))
    {
    yM[v] = u;
    return 1;
    }
    }
    else if(temp < slack)
    slack = temp;
    }
    return 0;
    }
    void EK()
    {
    int i, j, u;
    for(i = 0; i < N; i ++)
    {
    A[i] = 0;
    for(j = 0; j < N; j ++)
    if(G[i][j] > A[i])
    A[i] = G[i][j];
    }
    memset(B, 0, sizeof(B));
    memset(yM, -1, sizeof(yM));
    for(u = 0; u < N; u ++)
    for(;;)
    {
    memset(visx, 0, sizeof(visx));
    memset(visy, 0, sizeof(visy));
    slack = INF;
    if(searchpath(u))
    break;
    for(i = 0; i < N; i ++)
    {
    if(visx[i])
    A[i] -= slack;
    if(visy[i])
    B[i] += slack;
    }
    }
    }
    void printresult()
    {
    int i, res = 0;
    for(i = 0; i < N; i ++)
    res += MAX - G[yM[i]][i];
    printf("%d\n", res);
    }
    int main()
    {
    while(scanf("%d%d%d%d", &N, &M, &K, &P) == 4)
    {
    init();
    EK();
    printresult();
    }
    return 0;
    }


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  • 原文地址:https://www.cnblogs.com/staginner/p/2200835.html
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