UVALive 6908Electric Bike（dp）

题意:

给定N≤103个障碍物，和一辆车，变挡次数K≤10，车的电量E≤50，车有4挡，耗电为挡数，不同的挡提供不同的额外能量
能量不足以提供车当前挡前进的时候，车直接就报废了。问穿过所有的障碍物人的最少能量花费？

分析:

赤果果的dp
dp[i][k][e][l]:=第几个障碍物，剩余变挡次数，剩余能量，当前挡数，人的最小能量花费，由于状态1000∗10∗50∗4=2e6比较大，乘case数=100就T了
用队列优化一下就好了，类似于spfa的感觉

代码:

////  Created by TaoSama on 2015-12-21//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e3 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, k, e, ans;int h[N], f[N][11][51][4];bool in[N][11][51][4];int power[] = {0, 4, 8, 11};struct Sta {    int p, c, e, l; //position, chance, energy, level};void see(Sta u, Sta v) {    int &cur = f[u.p][u.c][u.e][u.l];    int &nxt = f[v.p][v.c][v.e][v.l];    printf("f[%d][%d][%d][%d] = %d → ", u.p, u.c, u.e, u.l, cur);    printf("f[%d][%d][%d][%d] = %d\n", v.p, v.c, v.e, v.l, nxt);}void spfa() {    queue<Sta> q;    memset(f, 0x3f, sizeof f);    memset(in, false, sizeof in);    f[0][k][e][0] = 0; in[0][k][e][0] = true;    q.push((Sta) {0, k, e, 0});    while(q.size()) {        Sta u = q.front(); q.pop();        in[u.p][u.c][u.e][u.l] = false;        int &cur = f[u.p][u.c][u.e][u.l];        if(u.p == n) {            ans = min(ans, cur);            continue;        }        if(u.c) {            for(int i = 0; i < 4; ++i) {                if(i == u.l || u.e < i) continue;                Sta v = u;                ++v.p; --v.c; v.e -= i; v.l = i;                int cost = max(0, h[v.p] - power[v.l]);                int &nxt = f[v.p][v.c][v.e][v.l];                if(nxt > cur + cost) {                    nxt = cur + cost;//                  see(u, v);                    if(!in[v.p][v.c][v.e][v.l]) {                        in[v.p][v.c][v.e][v.l] = true;                        q.push(v);                    }                }            }        }        Sta v = u; ++v.p;        if(v.e < v.l) v.c = v.e = v.l = 0; //lack energy, just set 0        else v.e -= v.l;        int cost = max(0, h[v.p] - power[v.l]);        int &nxt = f[v.p][v.c][v.e][v.l];        if(nxt > cur + cost) {            nxt = cur + cost;            if(!in[v.p][v.c][v.e][v.l]) {                in[v.p][v.c][v.e][v.l] = true;                q.push(v);            }        }    }}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    while(t--) {        scanf("%d%d%d", &n, &k, &e);        for(int i = 1; i <= n; ++i) scanf("%d", h + i);        ans = INF;        spfa();        static int kase = 0;        printf("Case #%d: %d\n", ++kase, ans);    }    return 0;}