UVa 12717 Fiasco (BFS模拟)

题意：给定一个错误代码，让你修改数据，使得它能够输出正确答案，错误代码是每次取最短的放入。

析：那么我们就可以模拟这个过程，然后修改每条边的权值，使得它能输出正确答案。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2500 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
vector<int> G[maxn];
vector<P> before;
bool vis[maxn];

int d[maxn][maxn];

void bfs(int s){
    queue<int> q;
    q.push(s);
    memset(vis, false, sizeof vis);

    int cnt = 0;
    while(!q.empty()){
        int u = q.front();  q.pop();
        if(vis[u])  continue;
        vis[u] = true;
        for(int i = 0; i < G[u].size(); ++i){
            int v = G[u][i];
            if(vis[v])  continue;
            d[u][v] = d[v][u] = ++cnt;
            q.push(v);
        }
    }
}

int main(){
    int T;  cin >> T;
    int s;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%d %d %d", &n, &m, &s);
        for(int i = 1; i <= n; ++i)  G[i].clear();

        int u, v, w;
        before.clear();
        for(int i = 0; i < m; ++i){
            scanf("%d %d %d", &u, &v, &w);
            G[u].push_back(v);
            G[v].push_back(u);
            before.push_back(P(u, v));
        }

        bfs(s);
        printf("Case %d:\n", kase);
        for(int i = 0; i < before.size(); ++i)
            printf("%d %d %d\n", before[i].first, before[i].second, d[before[i].first][before[i].second]);
    }
    return 0;
}