TopCoder SRM 642 Div.2 1000 --二分+BFS

题意： 给你一张图，N个点(0~N-1),m条边，国王要从0到N-1，国王携带一个值，当走到一条边权大于此值的边时，要么不走，要么提升该边的边权，提升k个单位花费k^2块钱，国王就带了B块钱，问能携带的最大值是多少。

解法:  二分此值，然后BFS跑遍每个点，记录到达每个点的最小花费Mincost，如果Mincost[N-1] <= B,则此值可行，往上再二分，否则往下二分。

比赛时候本来我的二分方法应该返回high的，结果返回low，怎么都过不了样例，比赛完才发现此处的问题。  真是太弱。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#define ll long long
using namespace std;

struct node
{
    int u;
    long long cost;
};

class TallShoes
{
public:
    long long mp[][],Mincost[];
    int N;
    bool bfs(int N,int S,int E,long long hei,long long B)
    {
        int i;
        Mincost[] = ;
        for(i=;i<N;i++)
            Mincost[i] = 10000000000000000LL;
        queue<node> que;
        node now;
        now.u = S;
        now.cost = ;
        que.push(now);
        while(!que.empty())
        {
            node tmp = que.front();
            que.pop();
            int u = tmp.u;
            long long cost = tmp.cost;
            for(i=;i<N;i++)
            {
                if(u == i) continue;
                if(mp[u][i] >= 10000000000000000LL) continue;
                if(mp[u][i] >= hei)
                {
                    if(Mincost[i] > cost)
                    {
                        Mincost[i] = cost;
                        now.u = i, now.cost = Mincost[i];
                        que.push(now);
                    }
                }
                else
                {
                    long long dif = hei-mp[u][i];
                    if(Mincost[i] > cost + dif*dif)
                    {
                        Mincost[i] = cost + dif*dif;
                        now.u = i, now.cost = Mincost[i];
                        que.push(now);
                    }
                }
            }
        }
        if(Mincost[E] <= B) return true;
        return false;
    }
    int maxHeight(int N, vector <int> X, vector <int> Y, vector <int> height, long long B)
    {
        for(int i=;i<N;i++)
        {
            for(int j=;j<N;j++)
                mp[i][j] = 10000000000000000LL;
            mp[i][i] = ;
        }
        for(int i=;i<X.size();i++)
            mp[X[i]][Y[i]] = mp[Y[i]][X[i]] = height[i];
        long long low = , high = 1000000000LL;
        while(low <= high)
        {
            long long mid = (low+high)/2LL;
            if(bfs(N,,N-,mid,B)) low = mid+;
            else                   high = mid-;
        }
        return high;
    }
};