BZOJ 3408: [Usaco2009 Oct]Heat Wave 热浪( 最短路 )

普通的最短路...dijkstra水过..

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#include<cstdio>

#include<algorithm>

#include<cstring>

#include<iostream>

#include<queue>

#define rep( i , n ) for( int i = 0 ; i < n ; i++ )

#define clr( x , c ) memset( x , c , sizeof( x ) )

using namespace std;

const int maxn = 2500 + 5;

const int maxm = 6200 + 5;

const int inf = 0x3f3f3f3f;

struct edge {

int to , dist;

edge* next;

};

edge* pt , EDGE[ maxm << 1 ];

edge* head[ maxn ];

void edge_init() {

pt = EDGE;

clr( head , 0 );

}

void add( int u , int v , int d ) {

pt -> to = v;

pt -> dist = d;

pt -> next = head[ u ];

head[ u ] = pt++;

}

#define add_edge( u , v , d ) add( u , v , d ) , add( v , u , d )

struct node {

int x , d;

bool operator < ( const node &rhs ) const {

return d > rhs.d;

}

};

int d[ maxn ];

priority_queue< node > Q;

void dijkstra( int S ) {

clr( d , inf );

d[ S ] = 0;

Q.push( ( node ) { S , 0 } );

while( ! Q.empty() ) {

node o = Q.top();

Q.pop();

int x = o.x , dist = o.d;

if( dist != d[ x ] ) continue;

for( edge* e = head[ x ] ; e ; e = e -> next ) {

int to = e -> to;

if( d[ to ] > dist + e -> dist ) {

d[ to ] = dist + e -> dist;

Q.push( ( node ) { to , d[ to ] } );

}

}

}

}

inline int read() {

char c = getchar();

int res = 0;

while( ! isdigit( c ) ) c = getchar();

while( isdigit( c ) ) {

res = res * 10 + c - '0';

c = getchar();

}

return res;

}

int main() {

    freopen( "test.in" , "r" , stdin );

    int n = read() , m = read() , s = read() - 1 , t = read() - 1;

    edge_init();

    while( m-- ) {

    int u = read() - 1 , v = read() - 1 , d = read();

    add_edge( u , v , d );

    }

    dijkstra( s );

    cout << d[ t ] << "\n";

return 0;

}

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3408: [Usaco2009 Oct]Heat Wave 热浪

Time Limit: 3 Sec  Memory Limit: 128 MB
Submit: 71  Solved: 59
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Description

Input

 第1行：4个由空格隔开的整数T，C，Ts，Te.

 第2到第C+1行：第i+l行描述第i条道路．有3个由空格隔开的整数Rs，Re，Ci.

Output

    一个单独的整数表示Ts到Te的最小费用．数据保证至少存在一条道路．

Sample Input

7 11 5 4
2 4 2
1 4 3
7 2 2
3 4 3
5 7 5
7 3 3
6 1 1
6 3 4
2 4 3
5 6 3
7 2 1

Sample Output

7

HINT

Source

Gold