zoj 1586 QS Network

最小生成树，刚刚学了Prim算法。

对每条边变的权值进行预处理，c[i][j] = c[i][j] + p[i] + p[j] 其中c[i][j]为输入的权值，p[i]，p[j]为连接这两个节点所需的费用。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
const int maxn = 1010;
int c[maxn][maxn];//邻接矩阵
int x[maxn], y[maxn]; //x数组表示i节点到集合的最短距离 y数组表示i节点和集合中哪个点是最短距离
int p[maxn];
int main()
{
    int sb;
    scanf("%d", &sb);
    while (sb--)
    {
        int n, m, i, j, u, v, cost;
        scanf("%d", &n);
        for (i = 1; i <= n; i++) scanf("%d", &p[i]);
        for (i = 1; i <= n; i++) for (j = 1; j <= n; j++) scanf("%d", &c[i][j]);
        for (i = 1; i <= n; i++) for (j = 1; j <= n; j++) c[i][j] = c[i][j] + p[i] + p[j];
        for (i = 1; i <= n; i++) x[i] = c[1][i], y[i] = 1;
        y[1] = -1;// y[i] == -1 表示i节点已经放入了集合
        int tot = 1;//已经有一个点放入了集合
        int ans = 0;
        while (1)
        {
            int mincost = 0x7FFFFFFF, v, flag = 0;
            for (i = 1; i <= n; i++) if (y[i] != -1 && x[i] < mincost) flag = 1, v = i, mincost = x[i];
            if (!flag) break;
            y[v] = -1; tot++; ans = ans + x[v];
            for (i = 1; i <= n; i++) if (y[i] != -1 && c[v][i] < x[i]) x[i] = c[v][i], y[i] = v;
        }
        if (tot == n)printf("%d
", ans);
        else printf("No Way!
");
    }
    return 0;
}