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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 longusing namespace std;struct node{    int u;    long long cost;};class TallShoes{public:    long long mp[55][55],Mincost[55];    int N;    bool bfs(int N,int S,int E,long long hei,long long B)    {        int i;        Mincost[0] = 0;        for(i=1;i<N;i++)            Mincost[i] = 10000000000000000LL;        queue<node> que;        node now;        now.u = S;        now.cost = 0;        que.push(now);        while(!que.empty())        {            node tmp = que.front();            que.pop();            int u = tmp.u;            long long cost = tmp.cost;            for(i=0;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=0;i<N;i++)        {            for(int j=0;j<N;j++)                mp[i][j] = 10000000000000000LL;            mp[i][i] = 0;        }        for(int i=0;i<X.size();i++)            mp[X[i]][Y[i]] = mp[Y[i]][X[i]] = height[i];        long long low = 0, high = 1000000000LL;        while(low <= high)        {            long long mid = (low+high)/2LL;            if(bfs(N,0,N-1,mid,B)) low = mid+1;            else                   high = mid-1;        }        return high;    }};
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TopCoder SRM 642 Div.2 1000 --二分+BFS