uVA 10600 ACM Contest and Blackout (prim求次小生成树)

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use标记当前边是否属于最小生成树。
d[i][j]代表从i->j的瓶颈路。
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int inf = 0x3f3f3f3f;int n , m , a , b , c;int map[105][105];int dist[105] , vis[105] , ans1 , ans2 , use[105][105] , p[205];int d[105][105];void prim(){    int tid , tmin;    ans1 = 0;    memset(vis , 0 , sizeof(vis));    memset(dist , 0 , sizeof(dist));    memset(p , -1 , sizeof(p));    for(int i = 1 ; i <= n ; i ++){        dist[i] = inf;    }    dist[1] = 0;    p[1] = -1;    for(int i = 0 ; i < n; i ++){        //printf("i = %d\n" , i);        tid = -1;        tmin = inf;        for(int j = 1 ; j <= n ; j ++){            if(vis[j] == 0 && tmin > dist[j]){                tmin = dist[j];                tid = j;            }        }        ans1 += tmin;        //printf("tmin = %d\n" , tmin);        vis[tid] = 1;        for(int j = 1 ; j <= n ; j ++){            if(vis[j] == 0 && dist[j] > map[tid][j]){                dist[j] = map[tid][j];                p[j] = tid;                //printf("j = %d pos = %d\n" , j , tid);            }        }    }    return ;}void dfs(int v){    for(int u = 1 ; u <= n ; u ++){        if(vis[u] == 0 && p[u] == v){            vis[u] = 1;            for(int x = 1 ; x <= n ; x ++){                if(vis[x] == 1 && x != u){                    d[x][u] = d[u][x] = max(d[x][v] , map[u][v]);                }            }            dfs(u);        }    }    return ;}void Second_MST(){    ans2 = inf;    for(int i = 1;i <= n; ++i){        for(int j = 1;j <= n; ++j){            if(use[i][j] == 1){                if(ans1 + map[i][j] - d[i][j] < ans2)                    ans2 = ans1 + map[i][j] - d[i][j];            }        }    }}int main(){    int t;    scanf("%d" , &t);    while(t --){        memset(map , inf , sizeof(map));        memset(use , 0 , sizeof(use));        scanf("%d%d" , &n , &m);        for(int i = 0 ; i < m ; i ++){            scanf("%d%d%d" , &a , &b , &c);            map[a][b] = c;            map[b][a] = c;            use[a][b] = use[b][a] = 1;        }        prim();        printf("%d " , ans1);        memset(vis , 0 , sizeof(vis));        memset(d , 0 , sizeof(d));        vis[1] = true;        dfs(1);        for(int i = 1 ; i <= n ; i ++){            if(p[i] != -1){                use[p[i]][i] = use[i][p[i]] = 0;            }        }        Second_MST();        printf("%d\n" , ans2);    }}