UVa 908 (RE-connectong computer sites )

這道題很唬人的，題目描述說了那麼長，開始還給出了很多不要的輸入，但是仔細讀題，第二個輸出數字，是指M+K條邊裡面選擇幾條邊，串連，是費用最小，其實就是把之前的費用加起來，然後再在M+K的情況下求一次最小產生樹

還有，資料很大，數組果斷不要用了，只能用vector表示的邊表，其中裡面還有的資料結構還是結構體

稠密圖，邊最多可達到10的11次冪，雖說，有些網友表示，測試資料其實並沒有那麼大，但prim明顯會更合適

代碼：

#include <cstdio>#include <cstring>#include <vector>using namespace std;const int N = 1000001;const int INF = 1000000000;struct edge { int to, c; };vector <edge> vec[N];int n, m, k, tre1;int prim() {    int mi, v, tre2 = 0;    vector<int> d;    d.push_back(INF);    d.push_back(0);    for ( int i = 2; i <= n; ++i ) d.push_back(INF);    bool vis[N];    memset(vis, 0, sizeof(vis));    for ( int u = 0; u < n; ++u ) {        mi = INF;        for ( int i = 1; i <= n; ++i ) if ( !vis[i] && d[i] < mi ) mi = d[i], v = i;        vis[v] = true;        tre2 += mi;        for ( int i = 0; i < vec[v].size(); ++i ) {            int x = vec[v][i].to, cost = vec[v][i].c;            if ( !vis[x] && d[x] > cost ) d[x] = cost;        }    }    return tre2;}int main(){    bool flag = false;    while ( scanf("%d", &n) != EOF ) {        if ( flag ) printf("\n");        flag = true;        tre1 = 0;        for ( int i = 1, x, y, z; i < n; tre1 += z, ++i ) scanf("%d%d%d", &x, &y, &z);        scanf("%d", &k);        for ( int i = 0; i <= n; ++i ) vec[i].clear();        for ( int i = 0; i < k; ++i ) {            int s, e, c;            edge x;            scanf("%d%d%d", &s, &e, &c);            x.to = e, x.c = c;            vec[s].push_back(x);            x.to = s, x.c = c;            vec[e].push_back(x);        }        scanf("%d", &m);        for ( int i = 0; i < m; ++i ) {            int s, e, c;            edge x;            scanf("%d%d%d", &s, &e, &c);            x.to = e; x.c = c;            vec[s].push_back(x);            x.to = s;            vec[e].push_back(x);        }        printf("%d\n", tre1);        printf("%d\n", prim());    }}