SSl 2404 上学#线性动态规划#

题目

某人至少要花多少钱才能到达学校。

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分析

spfa后，听取WA声一片，所以看了解题报告后发现是dp
（而且比深搜慢）
状态转移方程：$f[t+1][n]=min(f[i][j]+money[i])$

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dp代码

#include <cstdio>
#include <cctype>
#include <cstring>
using namespace std;
int n,t,m,x[51],y[51],l[51],r[51],w[51],f[100001][51];
int in(){
	int ans=0; char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=ans*10+c-48,c=getchar();
	return ans;
}
int main(){
	freopen("shaxu.in","r",stdin);
	freopen("shaxu.out","w",stdout);
	n=in(); t=in(); m=in(); memset(f,127/3,sizeof(f)); f[0][1]=0;
	for (int i=1;i<=m;i++) x[i]=1+in(),y[i]=1+in(),l[i]=in(),r[i]=l[i]+1+in(),w[i]=in();
	for (int i=0;i<=t;i++)
	for (int j=1;j<=n;j++){
		if (f[i][j]==707406378) continue;
	    for (int k=1;k<=m;k++)
	    if (x[k]==j&&l[k]==i&&r[k]<=t+1)
	    if (f[i][j]+w[k]<f[r[k]][y[k]]) f[r[k]][y[k]]=f[i][j]+w[k];
	    if (f[i+1][j]>f[i][j]) f[i+1][j]=f[i][j];
	}
	if (f[t+1][n]==707406378) return !puts("-1"); else return !printf("%d",f[t+1][n]);
}

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深搜代码

#include <cstdio>
#include <cctype>
#include <algorithm>
using namespace std;
unsigned short n,m,ls[51]; int mint=2147483647,t; bool v[51];
struct node{int x,y,must,last,w,next;}e[51];
int in(){
	int ans=0; char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=ans*10+c-48,c=getchar();
	return ans;
}
void dfs(int x,int cost,int now){
	if (x==n&&now<=t) {mint=min(mint,cost); return;}
	v[x]=1; int p=ls[x];
	while (p){
		if (!v[e[p].y]&&now<e[p].must)
		dfs(e[p].y,cost+e[p].w,e[p].must+e[p].last);
		p=e[p].next;
	}
	v[x]=0;
}
int main(){
	freopen("shaxu.in","r",stdin);
	freopen("shaxu.out","w",stdout);
	n=in(); t=in(); m=in();
	for (int i=1;i<=m;i++){
		e[i].x=in()+1; e[i].y=in()+1; e[i].must=in(); e[i].last=in();
		e[i].w=in(); e[i].next=ls[e[i].x]; ls[e[i].x]=i;
		if (e[i].x==1&&e[i].must==0) e[i].must++; 
	}
	dfs(1,0,0);
	if (mint==2147483647) return !puts("-1");
	return !printf("%d",mint);
}