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题意：给定一个有向图，你从1出发到n，走尽可能多的点，并且使总权值不大于t。
析：在比赛时，竟然看成有向图了，就想了好久，感觉dp，但是不会啊。。。如果是有向图就好做多了，枚举边，然后打印就好，dp[i][j] 表示，
经过 i 个结点，并且在 j的最小时间。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5000 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[maxn][maxn], p[maxn][maxn];
int G[maxn][3];
stack<int> stacks;
int main(){
int t;
while(scanf("%d %d %d", &n, &m, &t) == 3){
int x;
for(int i = 0; i < m*3; ++i){
scanf("%d", &x);
G[i/3][i%3] = x;
}
memset(dp, INF, sizeof dp);
dp[1][1] = 0;
memset(p, 0, sizeof p);
int ans = 0;
for(int i = 2; i <= n; ++i)
for(int j = 0; j < m; ++j){
int pre = G[j][0], last = G[j][1], w = G[j][2];
if(dp[i-1][pre] + w <= t && dp[i][last] > dp[i-1][pre] + w){
dp[i][last] = dp[i-1][pre] + w;
ans = last == n ? Max(ans, i) : ans;
p[i][last] = pre;
}
}
printf("%d\n", ans);
for(int i = n; ans; i = p[ans--][i]) stacks.push(i);
printf("%d", stacks.top()); stacks.pop();
while(!stacks.empty()) printf(" %d", stacks.top()), stacks.pop();
printf("\n");
}
return 0;
}
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5925546.html
