51nod 3 * problem

1640
题意：
一张无向图
在最小化最大边后求最大边权和
Slove：
sort
最小生成树
倒叙最大生成树

#include 
#include 
#include 
#include 
#include 
#include <string>

using namespace std;

#define LL long long

#define gc getchar()
inline int read() {int x = 0, f = 1; char c = gc; 
while(c < '0' || c > '9') {if(c == '-') f = -1; c = gc;}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x * f;}
inline LL read_LL() {LL x = 0; char c = gc; while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x;}
#undef gc

const int N = 1e5 + 10;

int fa[N];
int A[N << 1], U[N << 1], V[N << 1], W[N << 1];
int n, m;

bool Cmp(int a, int b) {return W[a] < W[b];}

int Get(int x) {return fa[x] == x ? x : fa[x] = Get(fa[x]);}

void Minst(int &R) {
    for(int i = 1; i <= n; i ++) fa[i] = i;
    int js = 0;
    for(int i = 1; i <= m; i ++) {
        int fu = Get(U[A[i]]), fv = Get(V[A[i]]);
        if(fu != fv) {
            fa[fu] = fv;
            js ++;
        }
        if(js == n - 1) {
            R = i;
            while(W[A[R + 1]] == W[A[i]]) R ++;
            return ;
        }
    }
}

inline long long Maxst(int R) {
    for(int i = 1; i <= n; i ++) fa[i] = i;
    int js = 0;
    long long ret = 0;
    for(int i = R; i >= 1; i --) {
        int fu = Get(U[A[i]]), fv = Get(V[A[i]]);
        if(fu != fv) {
            fa[fu] = fv;
            ret += W[A[i]];
            js ++;
        }
        if(js == n - 1) return ret;
    }
}

int main() {
    n = read(), m = read();
    for(int i = 1; i <= m; i ++) A[i] = i, U[i] = read(), V[i] = read(), W[i] = read();
    sort(A + 1, A + m + 1, Cmp);
    int R;
    Minst(R);
    cout << Maxst(R);
    return 0;
}

1649
由于 1 - n 之间一定存在一种直接相连的道路
判断哪种直接相连
跑另外一种的最短路

#include 
#include 
#include 
#include 
#include 
#include <string>

using namespace std;

#define LL long long

#define gc getchar()
inline int read() {int x = 0; char c = gc; while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x;}
inline LL read_LL() {LL x = 0; char c = gc; while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x;}
#undef gc

const int N = 410, oo = 999999999;

int Map[N][N], Bmap[N][N];
int n, m;

int main() {
    n = read(), m = read();
    for(int i = 1; i <= n; i ++) for(int j = 1; j <= n; j ++) Map[i][j] = oo;
    for(int i = 1; i <= n; i ++) Map[i][i] = 0;
    for(int i = 1; i <= n; i ++) for(int j = 1; j <= n; j ++) Bmap[i][j] = oo;
    for(int i = 1; i <= n; i ++) Bmap[i][i] = 0;
    for(int i = 1; i <= m; i ++) {
        int u = read(), v = read();
        Map[u][v] = Map[v][u] = 1;
    }
    if(Map[1][n] == 1) {
        for(int i = 1; i <= n; i ++) 
            for(int j = 1; j <= n; j ++) {
                if(i == j) continue;
                if(Map[i][j] == oo) Bmap[i][j] = 1;
            }
        for(int k = 1; k <= n; k ++)
            for(int i = 1; i <= n; i ++)
                for(int j = 1; j <= n; j ++)
                    Bmap[i][j] = min(Bmap[i][j], Bmap[i][k] + Bmap[k][j]);
        if(Bmap[1][n] == oo) cout << -1;
        else cout << Bmap[1][n];
    } else {
        for(int k = 1; k <= n; k ++)
            for(int i = 1; i <= n; i ++)
                for(int j = 1; j <= n; j ++)
                    Map[i][j] = min(Map[i][j], Map[i][k] + Map[k][j]);
        if(Map[1][n] == oo) cout << -1;
        else cout << Map[1][n];
    }
    return 0;
}

1535
图是树的充要条件
$m = n - 1$ && 图联通
由于题目无自环
所以不存在二元环
并且若 $m >= n - 1$
则图联通
此时若 $m = n$
那么就会存在且只存在一个三元环（或更大）
因此只需判断 $n = m$ 即可

if(n == m) cout << "FHTAGN!";
else cout << "NO";

 

转载于:https://www.cnblogs.com/shandongs1/p/9591918.html