HDU5669

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Problem:传送门

Portal

 原题目描述在最下面。

 给你n个点,m组边: a, b ,c ,d ,w.表示点[a,b]与点[c,d]有一条权值为w的无向边。求1-n的最短路。

 有k次把一条边权值变为0的魔法。


Solution:

 线段树优化建边。对于魔法就在求最短路的时候再加一维处理。

 dp[i][j]表示到点i用了j次魔法的最短路径长度。

\[dp[u][t] = min(dp[u][t],dp[v][t]+w);\\ dp[u][t+1] = min(dp[u][t+1],dp[v][t]);//t+1<=k \]


AC_Code:

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#define lson rt << 1
#define rson rt << 1 | 1
using namespace std;
typedef long long LL;

const LL INF = 0x3f3f3f3f;
const int MXN = 6e5 + 5;
const int MXE = 4e6 + 6;

int n, m, k;
int inde;
vector<int>vs, vt;
struct lp{
    int v, nex;
    int w;
    friend bool operator < (const lp &a,const lp &b) {
        return a.w > b.w;
    }
}cw[MXE], A, B;
int head[MXN], tot;
int dis[MXN][12];
int vis[MXN][12];
void add_edge(int u,int v,int w) {
    cw[++tot].v = v; cw[tot].w = w; cw[tot].nex = head[u];
    head[u] = tot;
}
int seg[MXN][2];
void build(int id,int l,int r,int rt) {
    seg[rt][id] = ++inde;
    if(l == r) {
        if(id) add_edge(seg[rt][id], l, 0);
        else add_edge(l, seg[rt][id], 0);
        return;
    }
    int mid = (l + r) >> 1;
    build(id, l, mid, lson); build(id, mid + 1, r, rson);
    if(id) {
        add_edge(seg[rt][id], seg[lson][id], 0); add_edge(seg[rt][id], seg[rson][id], 0);
    }else {
        add_edge(seg[lson][id], seg[rt][id], 0); add_edge(seg[rson][id], seg[rt][id], 0);
    }
}
void query(int id,int L,int R,int l,int r,int rt) {
    if(L <= l && r <= R) {
        if(id) vs.push_back(seg[rt][id]);
        else vt.push_back(seg[rt][id]);
        return;
    }
    int mid = (l + r) >> 1;
    if(L > mid) query(id, L, R, mid+1, r, rson);
    else if(R <= mid) query(id,L,R,l,mid,lson);
    else {
        query(id,L,mid,l,mid,lson); query(id,mid+1,R,mid+1,r,rson);
    }
}
void dij() {
    for(int i = 1; i <= inde; ++i) {
        for(int j = 0; j <= k; ++j) {
            dis[i][j] = INF;
            vis[i][j] = 0;
        }
    }
    dis[1][0] = 0;
    A.v = 1; A.w = 0; A.nex = 0;
    priority_queue<lp> Q;
    Q.push(A);
    while(!Q.empty()) {
        B = Q.top(); Q.pop();
        int u = B.v, t = B.nex;
        if(vis[u][t]) continue;
        vis[u][t] = 1;
        for(int i = head[u]; ~i; i = cw[i].nex) {
            int v = cw[i].v;
            if(dis[v][t] > dis[u][t] + cw[i].w) {
                dis[v][t] = dis[u][t] + cw[i].w;
                A.v = v; A.w = dis[v][t]; A.nex = t;
                Q.push(A);
            }
            if(t < k && dis[v][t+1] > dis[u][t]) {
                dis[v][t+1] = dis[u][t];
                A.v = v; A.w = dis[v][t+1]; A.nex = t+1;
                Q.push(A);
            }
        }
    }
    int ans = INF;
    for(int i = 0; i <= k; ++i) ans = min(ans, dis[n][i]);
    if(ans >= INF) printf("CreationAugust is a sb!\n");
    else printf("%d\n", ans);
}
int main() {
    int tim;
    scanf("%d", &tim);
    while(tim --) {
        scanf("%d%d%d", &n, &m, &k);
        int a, b, c, d, w;
        memset(head, -1, sizeof(head));
        tot = -1;
        inde = n;
        build(1, 1, n, 1); build(0, 1, n, 1);
        for(int i = 0, ta, tb; i < m; ++i) {
            scanf("%d%d%d%d%d", &a, &b, &c, &d, &w);
            vs.clear(); vt.clear();
            query(1,c,d,1,n,1); query(0,a,b,1,n,1);
            ta = vs.size(); tb = vt.size();
            for(int i = 0; i < tb; ++i) {
                for(int j = 0; j < ta; ++j) {
                    add_edge(vt[i], vs[j], w);
                }
            }
            vs.clear(); vt.clear();
            query(1,a,b,1,n,1); query(0,c,d,1,n,1);
            a = vs.size(); b = vt.size();
            for(int i = 0; i < b; ++i) {
                for(int j = 0; j < a; ++j) {
                    add_edge(vt[i], vs[j], w);
                }
            }
        }
        dij();
    }
    return 0;
}

Problem Description:

posted @ 2018-10-09 18:11  Cwolf9  阅读(104)  评论(0编辑  收藏  举报