HDU-2962-Trucking

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ACM模版

描述

题解

在满足限制条件下，求最大高度情况下的最短路。
SPFA+二分即可。

代码

#include <cstdio>#include <cstring>#include <vector>#include <queue>#define MEM(a, v) memset (a, v, sizeof(a))using namespace std;const int INF = 0x3f3f3f3f;const int MAXN = 1010;struct edge{    int to;    int hei, dis;};bool used[MAXN];int c, r;int dist[MAXN];vector<edge> map[MAXN];int spfa(int beg, int end, int lim){    queue<int> q;    edge e;    MEM(dist, 0x3f);    MEM(used, 0);    q.push(beg);    used[beg] = 1;    dist[beg] = 0;    while (!q.empty())    {        int t = q.front();        q.pop();        for (int i = 0; i < map[t].size(); ++i)        {            e = map[t][i];            if (e.hei >= lim && dist[t] + e.dis < dist[e.to])            {                dist[e.to] = dist[t] + e.dis;                if (!used[e.to])                {                    used[e.to] = 1;                    q.push(e.to);                }            }        }        used[t] = 0;    }    return dist[end];}void init (){    for (int i = 1; i <= c; ++i)    {        map[i].clear();    }}int main (){    int key = 0;    while (scanf("%d%d", &c, &r), c | r)    {        edge e;        init();        int x, y, h = 0, d;        for (int i = 1; i <= r; ++i)        {            scanf("%d%d%d%d", &x, &y, &h, &d);            h = (h == -1 ? INF : h);            e.to = y;            e.hei = h;            e.dis = d;            map[x].push_back(e);            e.to = x;            map[y].push_back(e);        }        int low = 1, mid, high;        scanf("%d%d%d", &x, &y, &high);        //  二分查找，暴力枚举所有高度        int res = INF, ans = INF;        while (low <= high)        {            mid = (low + high) / 2;            res = spfa(x, y, mid);            if (INF == res)            {                high = mid - 1;            }            else            {                low = mid + 1;                ans = res;                h = mid;            }        }        if (key)        {            putchar('\n');        }        printf("Case %d:\n", ++key);        if (ans != INF)        {            printf("maximum height = %d\nlength of shortest route = %d\n", h, ans);        }        else        {            printf ("cannot reach destination\n");        }    }    return 0 ;}

参考

《最短路》
《二分查找》

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