The 37th ACM/ICPC Asia Regional Tianjin Site Online Contest - G Travel
这道题比赛的时候我做了2个多钟头..最后还是没过..各种都觉得和谐...刚才才发现是因为一个分号..在输入里的一个if语句后加了分号..使得这个判断失效了..囧暴了...去掉分号..果断AC...
首先当然是Floyd了..得到两两间的最短距离...
首先当然是Floyd了..得到两两间的最短距离...
方法一: 也就是我比赛的时候这么写的...用DP来解旅行商问题...15个城市可以用15位二进制数表示.最大32767...用dp[k][i]表示在k状态下(0~32767)最后到达的是i城市的最大剩余money..
Program:
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<map>
#include<queue>
#include<stack>
#define ll long long
#define oo 1000000000
#define pi acos(-1)
using namespace std;
struct node
{
int t,c,d;
}p[17];
int arc[105][105],n,m,money,goal,dp[70000][17];
void Floyd()
{
int k,i,j;
for (k=1;k<=n;k++)
for (i=1;i<=n;i++)
for (j=1;j<=n;j++)
if (arc[i][k]!=-1 && arc[k][j]!=-1)
if (arc[i][j]==-1 || arc[i][j]>arc[i][k]+arc[k][j])
arc[i][j]=arc[i][k]+arc[k][j];
return;
}
bool EXE_DP()
{
int i,j,ans,k,h,x,w,data;
memset(dp,-1,sizeof(dp));
p[0].t=1; p[0].c=p[0].d=0;
w=1;
dp[0][0]=money;
for (k=1;k<=goal;k++)
{
h=k;
x=1;
i=0;
while (h)
{
i++;
if (h%2)
{
w=k-x;
for (j=0;j<=n;j++)
if (dp[w][j]!=-1 && arc[p[j].t][p[i].t]!=-1)
if (dp[w][j]-arc[p[j].t][p[i].t]>=p[i].d)
{
data=dp[w][j]-arc[p[j].t][p[i].t]+p[i].c-p[i].d;
if (dp[k][i]==-1 || dp[k][i]<data) dp[k][i]=data;
}
}
x*=2;
h/=2;
}
}
for (i=1;i<=n;i++)
if (arc[p[i].t][1]!=-1 && dp[goal][i]!=-1 && dp[goal][i]>=arc[p[i].t][1])
return true;
return false;
}
int main()
{
// freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);
int t,x,y,h,w;
scanf("%d",&t);
while (t--)
{
memset(arc,-1,sizeof(arc));
scanf("%d%d%d",&n,&m,&money);
while (m--)
{
scanf("%d%d",&x,&y);
scanf("%d",&w);
if (arc[y][x]==-1 || w<arc[y][x])
arc[x][y]=arc[y][x]=w;
}
for (x=1;x<=n;x++) arc[x][x]=0;
Floyd();
scanf("%d",&h);
goal=1;
for (x=1;x<=h;x++)
{
scanf("%d%d%d",&p[x].t,&p[x].c,&p[x].d);
goal*=2;
}
n=h;
goal--;
if (EXE_DP()) printf("YES\n");
else printf("NO\n");
}
return 0;
}
Program:
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<map>
#include<queue>
#include<stack>
#define ll long long
#define oo 10000000
#define pi acos(-1)
using namespace std;
struct node
{
int t,c,d;
}p[17];
int arc[105][105],n,m,h;
bool used[17];
void Floyd()
{
int k,i,j;
for (k=1;k<=n;k++)
for (i=1;i<=n;i++)
for (j=1;j<=n;j++)
if (arc[i][j]-arc[i][k]>arc[k][j])
arc[i][j]=arc[i][k]+arc[k][j];
return;
}
bool DFS(int money,int k,int now)
{
if (k==h)
{
if (money>=arc[p[now].t][1]) return true;
return false;
}
used[now]=true;
for (int i=1;i<=h;i++)
if (!used[i] && money-arc[p[now].t][p[i].t]>=p[i].d)
if (DFS(money-arc[p[now].t][p[i].t]-p[i].d+p[i].c,k+1,i)) return true;
used[now]=false;
return false;
}
int main()
{
freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);
int t,x,y,w,money;
scanf("%d",&t);
while (t--)
{
scanf("%d%d%d",&n,&m,&money);
for (y=0;y<=n;y++)
for (x=0;x<n;x++)
arc[y][x]=arc[x][y]=oo;
while (m--)
{
scanf("%d%d%d",&x,&y,&w);
if (w<arc[y][x])
arc[x][y]=arc[y][x]=w;
}
for (x=1;x<=n;x++) arc[x][x]=0;
Floyd();
scanf("%d",&h);
for (x=1;x<=h;x++)
scanf("%d%d%d",&p[x].t,&p[x].c,&p[x].d);
p[0].t=1; p[0].c=p[0].d=0;
memset(used,false,sizeof(used));
if (DFS(money,0,0)) printf("YES\n");
else printf("NO\n");
}
return 0;
}