• POJ


    POJ - 1511 题目链接

    一个非常玄学的操作

    同样的代码,我用c++就直接wa,用G++成功ac,这是为什么???

    说题目思路吧

    这显然是一题建立反向边的题,n稍微大一点到了1e6的级别,所以朴素版的Dijkstra是肯定过不了的,考虑用堆优化的1e6log(1e6)稳过,这道题目限时8000ms。
    之前也有一道建立方向边的题目,懒得打代码,直接抄下来,改了一改直接用了

    还有一个要注意的就是这道题目要用long long,1e6 * 1e6嘛,不在int的范围内。

    代码

    //Powered by CK
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    using namespace std;
    typedef long long ll;
    typedef pair<ll, ll> PII;
    const ll INF = 0x3f3f3f3f3f3f3f3f;
    const int N1 = 1e6 + 10, N2 = 1e6 + 10;
    ll head1[N1], to1[N2], nex1[N2], value1[N2], visit1[N1], dis1[N1], cnt1 = 1;
    ll head2[N1], to2[N2], nex2[N2], value2[N2], visit2[N1], dis2[N2], cnt2 = 1;
    int n, m, s;
    struct cmp {
        bool operator () (const PII & a, const PII & b) const {
            return a.second > b.second;
        }
    };
    void add1(ll x, ll y, ll w) {
        to1[cnt1] = y;
        value1[cnt1]  = w;
        nex1[cnt1] = head1[x];
        head1[x] = cnt1++;
    }
    void add2(ll x, ll y, ll w) {
        to2[cnt2] = y;
        value2[cnt2] = w;
        nex2[cnt2] = head2[x];
        head2[x] = cnt2++;
    }
    void Dijkstra1() {
        priority_queue<PII, vector<PII>, cmp> q;
        for(int i = 1; i <= n; i++) dis1[i] = INF;
        dis1[1] = 0;
        q.push(make_pair(1, 0));
        while(!q.empty()) {
            ll temp = q.top().first;
            q.pop();
            if(visit1[temp])    continue;
            visit1[temp] = 1;
            for(ll i = head1[temp]; i; i = nex1[i])
                if(dis1[to1[i]] > dis1[temp] + value1[i]) {
                    dis1[to1[i]] = dis1[temp] + value1[i];
                    q.push(make_pair(to1[i], dis1[to1[i]]));
                }
        }
    }
    void Dijkstra2() {
        priority_queue<PII, vector<PII>, cmp> q;
        for(int i = 1; i <= n; i++) dis2[i] = INF;
        dis2[1] = 0;
        q.push(make_pair(1, 0));
        while(!q.empty()) {
            ll temp = q.top().first;
            q.pop();
            if(visit2[temp])    continue;
            visit2[temp] = 1;
            for(ll i = head2[temp]; i; i = nex2[i])
                if(dis2[to2[i]] > dis2[temp] + value2[i]) {
                    dis2[to2[i]] = dis2[temp] + value2[i];
                    q.push(make_pair(to2[i], dis2[to2[i]]));
                }
        }
    }
    int main() {
        // freopen("in.txt", "r", stdin);
        ll x, y, w, t;
        scanf("%lld", &t);
        while(t--) {
            scanf("%lld %lld", &n, &m);
            memset(visit1, 0, sizeof visit1);
            memset(visit2, 0, sizeof visit2);
            memset(head1, 0, sizeof head1);
            memset(head2, 0, sizeof head2);
            cnt1 = cnt2 = 1;
            for(int i = 0; i < m; i++) {
                scanf("%lld %lld %lld", &x, &y, &w);
                add1(x, y, w);
                add2(y, x, w);
            }
            Dijkstra1();
            Dijkstra2();
            ll ans = 0;
            for(int i = 1; i <= n; i++)
                ans += dis1[i] + dis2[i];
            printf("%lld
    ", ans);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/lifehappy/p/12625767.html
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