HDU5669
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Catalog
Problem:传送门
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:
ACMer,无怨无悔